Gujarat Board GSEB Textbook Solutions Class 11 Maths Chapter 9 Sequences and Series Ex 9.2 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 11 Maths Chapter 9 Sequences and Series Ex 9.2

Question 1.

Find the sum of odd integers from 1 to 2001.

Solution:

We have to find the sum 1 + 3 + 5 + ………….. + 2001.

Let the nth term be the last term.

âˆ´ a + (n – 1)d = 2001.

Here, a = 1, d = 2.

âˆ´ 1 + [n – 1].2 = 2001 or 2[n – 1] = 2000 – 1 = 2000

âˆ´ n – 1 = 2000, n = 1001

âˆ´ Sum = \(\frac{n}{2}\) [2a + (n – 1)d]

= \(\frac{1001}{2}\)[2.1 + (1001 – 1) Ã— 2]

= \(\frac{1001}{2}\) Ã— 2[1 + 100 – 1] = 1001 Ã— 1001

= 1002001.

Question 2.

Find the sum of all natural numbers lying between 100 and 1000 which are multiples of 5?

Solution:

We have to find the sum 105 + 110 + 115 + … 995

Let 995 = nth term.

âˆ´ a + (n – 1)d = 995.

n = 199 – 20 = 179

âˆ´ 105 + 110 + 115 + ……….. + 995 = \(\frac{n}{2}\)[2a + (n – 1)d]

= \(\frac{179}{2}\) [2 Ã— 105 + (179 – 1) Ã— 5]

= \(\frac{179}{2}\) [2 Ã— 105 + 5 Ã— 178]

= 179(105 + 5 Ã— 89)

= 179(105 + 445) = 179 Ã— 550

= 98450.

Question 3.

In an A.P., the first term is 2 and the sum of first five terms is one-fourth of the sum of next five terms. Show that its 20th term is – 112.

Solution:

Let d be the common difference.

âˆ´ Sum of first 5 terms = \(\frac{n}{2}\) [(2a + (n – 1))d]

= \(\frac{5}{2}\) [2 Ã— 2 + (5 – 1)d]

= \(\frac{5}{2}\)[4 + 4d] = 5(2 + 2d).

6^{th}term = a + (6 – 1)d = a + 5d.

âˆ´ (a + 5d) + (a + 6d) + ………….. to 5 terms

= \(\frac{5}{2}\) [2(a + 5d) + (5 – 1)d]

= \(\frac{5}{2}\) [2a + 10d + 4d]

= \(\frac{5}{2}\) [2a + 14d] = 5(a + 7d)

= 5(2 + 7d)

Sum of first five terms = \(\frac{1}{4}\) Ã— sum of next 5 terms

5(2 + 2d) = \(\frac{1}{4}\) Ã— 5(2 + 7d)

or 4(2 + 2d) = 2 + 7d

âˆ´ d = – 6

âˆ´ T_{20} = a + (20 – 1)d = 2 + 19 Ã— (- 6)

= 2 + (- 114) = – 112.

4. How many terms of the A.P. – 6, – \(\frac{11}{2}\), – 5, ……………… are needed to give the sum – 25?

Solution:

Here, a = – 6, d = – \(\frac{11}{2}\) – (- 6) = \(\frac{- 11}{2}\), – 5, …………….. are needed to give the sum – 25?

Solution:

Here, a = – 6, d = – \(\frac{11}{2}\) – (- 6) = \(\frac{- 11}{2}\) + 6 = \(\frac{- 11+12}{2}\) = \(\frac{1}{2}\).

Let n be the number of terms, so that

S_{n} = – 25.

Since S_{n} = \(\frac{n}{2}\)[2a + (n – 1)d], we have:

â‡’ – 100 = n(n – 25)

â‡’ n^{2} – 25n + 100 = 0

â‡’ (n – 5)(n – 20) = 0

â‡’ n = 5 or n = 20.

Both values of n give the required sum.

Question 5.

In an A.P., if the pth term is \(\frac{1}{q}\) and the qth term is \(\frac{1}{p}\), prove that the sum of first pq terms is \(\frac{1}{2}\)(pq + 1), where p â‰ q?

Solution:

Let a be the first term and d be the common difference of Â£he A.P. Then,

Putting d = \(\frac{1}{pq}\) in (1), we get

Question 6.

If the sum of a certain number of terms of the A.P; 25, 22, 19, ………………. is 116. Find the last term?

Solution:

a = 25, d = 22 – 25 = – 3. Let n be the number of terms.

Sum = 116.

Now, sum = \(\frac{n}{2}\) [2a + (n – 1)d]

âˆ´ 116 = \(\frac{n}{2}\) [50 + (n – 1)(- 3)]

or 232 = n[50 – 3n + 3] = n[53 – 3n]

= – 3n^{2} + 53n.

â‡’ 3n^{2} – 53n + 232 = 0.

â‡’ (n – 8)(3n – 29) = 0.

âˆ´ n = 8 or \(\frac{29}{3}\). But n â‰ \(\frac{29}{3}\).

âˆ´ n = 8.

Now, T_{8} = a + (8 – 1)d = 25 + 7 Ã— (- 3)

= 25 – 21 = 4.

âˆ´ Last term = 4.

Question 7.

Find the sum to n terms of an A.P., whose kth term is 5k + 1.

Solution:

T_{k} = 5k + 1.

Putting k = 1 and 2, we get

T_{1} = 5 Ã— 1 + 1 = 5 + 1 = 6

and T_{2} = 5 Ã— 2 + 1 = 10 + 1 = 11.

âˆ´ d = T_{2} – T_{1} = 11 – 6 = 5.

Here, a = 6, d = 5.

Sum of n terms = \(\frac{n}{2}\)[2a + (n – 1)d].

= \(\frac{n}{2}\)[2 Ã— 6 + (n – 1) Ã— 5]

= \(\frac{n}{2}\)[12 + 5n – 5]

= \(\frac{n(5n+7)}{2}\).

Question 8.

If the sum to n terms of an A.P. is (pn + qn^{2}), where p and q are constants, find the common difference?

Solution:

Let S_{n} be the sum to n terms.

âˆ´ S_{n} = pn + qn^{2}

Putting n = 1 and 2, we get

S_{1} = T_{1} = p.1 + q.1^{2} = p + q

and S_{2} = T_{1} + T_{2} = p.2 + q.2^{2} = 2p + 4q.

T_{2} = (T_{1} + T_{2}) – T_{1} = S_{2} – S_{1}

= [(2p + 4q)] – (p + q)]

= p + 3q.

âˆ´ d = T_{2} – T_{1} = (p + 3q) – (p + q) = 2q.

âˆ´ Common differences of the A.P. is 2q.

Question 9.

The sums of first n terms of two arithmetic progressions are in the ratio (5n + 4) : (9n + 6). Find the ratio of their 18th terms?

Solution:

Let a_{1} and d_{1} be the first term and common difference of the first A.P. and a_{2} and d_{2} be the first term and common difference of the second A.P.

If S_{1} and S_{2} be their sums respectively, then

Multiplying numerator and denominator by 2, we get

Question 10.

If the sum of first p terms of an A.P. is equal to the sum of first q terms, then find the sum of first (p + q) terms?

Solution:

Let a be the first term and d be the common difference of A.P.

âˆ´ Sum of first p terms = \(\frac{p}{2}\)[2a + (p – 1)d] ……………. (1)

and sum of first q terms = \(\frac{q}{2}\)[2a + (q – 1)d] …………….. (2)

Equating (1) and (2), we get

\(\frac{p}{2}\)[(2a + (p – 1)d] = \(\frac{q}{2}\)[2a + (q – 1)d].

Transposing the terms of R.H.S. to L.H.S., we get

2a(p – q) + [p(p – 1) – q(q – 1)d] = 0

or 2a(p – q) + [p^{2} – q^{2} – (p – q)d] = 0

or 2a(p – q) + (p – q)[(p + q) – d] = 0

or (p – q)[2a + (p + q – 1)d] = 0

â‡’ 2a + (p + q – 1)d = 0

Sum of first (p + q) terms = \(\frac{p+q}{2}\)[2a + (p + q – 1)d]

= \(\frac{p+q}{2}\) Ã— 0 = 0

[âˆµ 2a + (p + q – 1)d = 0 from (3)]

Question 11.

Sum of the first p, q and r terms of an A.P. are a, b and c respectively. Prove that \(\frac{a}{p}\)(q – r) + \(\frac{b}{q}\)(r – p) + \(\frac{c}{r}\)(p – q) = 0.

Solution:

Let A be the first term and D be the common difference of A.P.

S_{p} = a.

Multiplying (1) by (q – r), (2) by (r – p), (3) by (p – q) and adding we get

A(q – r + r – p + p – q) + \(\frac{1}{2}\)[(p – 1)(q – r)(r – q) + (r – 1)(p – q)]D

= \(\frac{a}{p}\)(q – r) + \(\frac{b}{q}\)(r – p) + \(\frac{c}{r}\)(p – r)

â‡’ A(0) + \(\frac{1}{2}\)(0)D = \(\frac{a}{p}\)(q – r) + \(\frac{b}{q}\)(r – p) + \(\frac{c}{r}\)(p – r).

Hence, \(\frac{a}{p}\)(q – r) + \(\frac{b}{q}\)(r – p) + \(\frac{c}{r}\)(p – q) = 0.

Question 12.

The ratio of the sums of first m and the terms of an A.P. is m^{2} : n^{2}. Show that the ratio of mth and nth terms is 2m – 1 : 2n – 1.

Solution:

Let a be the first term and d be the common difference of A.P.

i.e; We are to find the value of \(\frac{a+(m – 1)d}{a+(n – 1)d}\) = \(\frac{2a+(2m-2)d}{2a+(2n-2)d}\)

L.H.S. of (1) will be identical with (2), if (m – 1) replaces (2m – 2) and (n – 1) replaces (2n – 2). So, replacing m by 2m – 1 and n by 2n – 1 on both sides of (1), we get

\(\frac{2a+(2m-2)d}{2a+(2n-2)d}\) = \(\frac{(2m-1)}{(2n-1)}\).

Question 13.

If the sum of first re terms of an A.P. is 3n^{2} + 5n and its mth term is 164, find the value of m?

Solution:

Let the sum of first re terms be denoted by S_{p}.

âˆ´ S_{n} = 3n^{2} + 5n.

Put n = 1 and 2.

T_{1} = S_{1} = 3.1^{2} + 5.1 = 3 + 5 = 8.

S_{2} = T_{1} + T_{2} = 3.2^{2} + 5.2 = 12 + 10 = 22.

âˆ´ T_{2} = S_{2} – S_{1} = 22 – 8 = 14.

âˆ´ Common difference d = T_{2} – T_{1} = 14 – 8 = 6.

a = 8, d = 6.

mth term = a + (m – 1)d = 164.

or 8 + (m – 1)6 = 164.

â‡’ 6m + 2 = 164 or 6m = 164 – 2 = 162.

âˆ´ m = \(\frac{162}{6}\) = 27.

Question 14.

Insert five numbers between 8 and 26 such that the resulting sequence is an A.P.

Solution:

Let the five numbers be denoted by T_{2}, T_{3}, T_{4}, T_{5} and T_{6}.

âˆ´ 8, T_{2}, T_{3}, T_{4}, T_{5}, T_{6} and 26 are in A.P.

26 is the 7th term. If d is the common difference,

then T_{7} = a + (7 – 1)d.

â‡’ 26 = 8 + 6d â‡’ 6d = 26 – 8 = 18

âˆ´ d = 3.

So, T_{2} = a + d = 8 + 3 = 11,

T_{3} = a + 2d = 8 + 16 = 14,

T_{4} = a + 3d = 8 + 9 = 17,

T_{5} = a + 4d = 8 + 12 = 20,

and T_{6} = a + 5d = 8 + 15 = 23.

Required five numbers are 11, 14, 17, 20 and 23.

Question 15.

If \(\frac{a^{n}+b^{n}}{a^{n-1}+b^{n-1}}\) is the A.M. between a and b, then find the value of n?

Solution:

A.M. between a and b = \(\frac{a+b}{2}\)

âˆ´ \(\frac{a^{n}+b^{n}}{a^{n-1}+b^{n-1}}\) = \(\frac{a+b}{2}\)

â‡’ 2a^{n} + 2b^{n} = a^{n} + ab^{n-1} + a^{n-1}b + b^{n}

â‡’ a^{n} – a^{n-1}b – ab^{n-1} + b^{n} = 0

â‡’ a^{n-1}(a – b) – b^{n-1}(a – b) = 0

â‡’ (a – b)(a^{n-1} – b^{n-1}) = 0 [âˆµa â‰ 0]

â‡’ a^{n-1} – b^{n-1} = 0

â‡’ a^{n-1} = b^{n-1}.

â‡’ \(\frac{a}{b}\)^{n-1} = 1 = (\(\frac{a}{b}\))^{0}

â‡’ n – 1 = 0 â‡’ n = 1.

Question 16.

Between 1 and 31, m numbers have been inserted in such a way that the resulting sequence is an A.P. and the ratio of 7th and (m – 1)th numbers is 5:9. Find the value of m?

Solution:

Let A_{1}, A_{2}, ……………….. A_{m} be the m arithmetic means each inserted between 1 and 31.

Total number of terms = m + 2.

If d is the common difference, then

1 + (m + 2 – 1 )d = 31.

â‡’ (m + 1)d – 30.

â‡’ d = \(\frac{30}{m+1}\).

So, A_{7} = T_{8} = a + 7d

= 1 + 7 Ã— \(\frac{30}{m+1}\)

= \(\frac{m+1+210}{m+1}\) = \(\frac{m+211}{m+1}\)

Also, A_{m-1} = T_{m} = a + (m – 1)d

= 1 + (m – 1) Ã— \(\frac{30}{m+1}\)

= \(\frac{m+1+30m-30}{m+1}\) = \(\frac{31m-29}{m+1}\)

Now, \(\frac{A_{7}}{A_{m-1}}\) = \(\frac{m+211}{31m-29}\) = \(\frac{5}{9}\).

â‡’ 9m + 1899 = 155m – 145

â‡’ 146m = 2044

â‡’ m = \(\frac{2044}{146}\) = 14

Hence, m = 14.

Question 17.

A man starts repaying a loan as first instalment of â‚¹ 100. If he increases the instalment by â‚¹ 5 every month, what amount he will pay in the 30th instalment?

Solution:

The instalments form an A.P., whose first term is â‚¹ 100 and the common difference is â‚¹ 5.

âˆ´ 30th instalment = T_{30} = a + (30 – 1)d

= 100 + 29 Ã— 5

= 100 + 143

= 245.

Thus, the 30th instalment is â‚¹ 245.

Question 18.

The difference between any two consecutive interior angles of a polygon is 5Â°. If the smallest angle is 120Â°, find the number of sides of the polygon?

Solution:

The angles of a polygon of n sides form an A.P., whose first term is 120Â° and common difference is 5Â°.

The sum of interior angles (in degrees)

= \(\frac{n}{2}\) [2a + (n – 1)d]

= \(\frac{n}{2}\) [2 Ã— 120 + (n – 1) Ã— 5]

= \(\frac{n}{2}\) [240 + 5n – 5] = \(\frac{n}{2}\)(235 + 5n).

Also, the sum of interior angles (in degrees)

= 180 Ã— n – 360.

âˆ´ \(\frac{n}{2}\)(235 + 5n) = 180n – 360

Multiplying by \(\frac{2}{5}\), we get

n(47 + n) = 2(36n – 72)

= 72n – 144

or n^{2} + (47 – 72)n + 144 = 0.

â‡’ n^{2} – 25n + 144 = 0

â‡’ (n – 16)(n – 9) = 0

Now, n â‰ 16

âˆ´ n = 9.