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# Exercise 9.3 (Available) Class 11 maths 9. Sequences and Series - ncert solutions - Toppers Study

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## Exercise 9.3 (Available) Class 11 maths 9. Sequences and Series - ncert solutions - Toppers Study

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### Exercise 9.3 (Available) class 11 Mathematics Chapter 9. Sequences and Series

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## 9. Sequences and Series

### | Exercise 9.3 (Available) |

## Exercise 9.3 (Available) Class 11 maths 9. Sequences and Series - ncert solutions - Toppers Study

**Q13. How many terms of G.P. 3, 3 ^{2}, 3^{3}, …. are needed to give the sum 120?**

**Q14. The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128. Determine the first term, the common ratio and the sum to n terms of the G.P.**

**Solution: **

**Q15. Given a G.P. with a = 729 and 7th term 64, determine S_{7}.**

**Solution: **

**Q16. Find a G.P. for which sum of the first two terms is – 4 and the fifth term is 4 times the third term.**

**Solution: **

S_{2} = - 4,

T_{5} = 4(T_{3})

**Q17. If the 4th, 10th and 16th terms of a G.P. are x, y and z, respectively. Prove that x, y, z are in G.P.**

**Solution: **

T_{4} = ar^{3} = x ------------------ (I)

T_{10} = ar^{9} = y ---------------(II)

T_{16 }= ar^{15 } = z ---------------(III)

If T_{4}, T_{10} and T_{16} are in G.P then x, y and z also will be in G.P

**Q18. Find the sum to n terms of the sequence, 8, 88, 888, 8888… .**

**Solution: **

Let S is the sum of n terms of series;

∴ S_{n } = 8 + 88 + 888 + 8888 + ………….. to the n term

= 8(1 + 11 + 111 + 1111 + ……….. )

**Q20. Show that the products of the corresponding terms of the sequences a, ar, ar^{2}, …ar^{n }^{– 1} and A, AR, AR^{2}, … AR^{n – }^{1} form a G.P, and find the common ratio.**

**Solution: **

Product of sequence = a.A, ar.AR, ar^{2}.AR^{2} ………….. ar^{n -1}.AR^{n -1}

Common ratio :

**Q21. Find four numbers forming a geometric progression in which the third term is greater than the first term by 9, and the second term is greater than the 4th by 18.**

**Solution: **

**Q22. If the p^{th}, q^{th} and r^{th} terms of a G.P. are a, b and c, respectively. Prove that**

*a ^{q – r} b^{r – p} c^{p – q} *= 1.

**Solution:**

Let first term be A and common ratio be R.

T_{p} = AR^{p – 1 }= a -------------- (I)

T_{q} = AR^{q – 1 }= b -------------- (II)

T_{r} = AR^{r – 1 }= c --------------- (III)

a^{q – r }. b^{r – p }. c^{p – q }= (AR^{p – 1 })^{q – r }. (AR^{q – 1})^{r – p } . (AR^{r – 1})^{p – q }

= AR^{(p – 1)(q – r )}.AR^{(q – 1)(r – p) }. AR^{(r – 1)(p – q)}

= AR^{(p – 1)(q – r ) + (q – 1)(r – p) + (r – 1)(p – q)}

= AR^{(pq – pr – q + r + qr –} ^{pq – r + p} ^{+ pr – qr – p + q)}

= (AR)^{0}

= 1

**Q23. If the first and the n ^{th} term of a G.P. are a and b, respectively, and if P is the product of n terms, prove that P^{2} = (ab)^{n}.**

**Solution: **

**Q24.** **Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from (n + 1)^{th} to (2n)^{th }term is 1/r^{n}.**

**Solution: **

**Q25. If a, b, c and d are in G.P. show that**

** ( a^{2} + b^{2} + c^{2}) (b^{2} + c^{2} + d^{2}) = (ab + bc + cd)^{2}.**

**Solution: **

a, b, c, d are in G.P. Therefore,

bc = ad ……………………... (1)

b^{2} = ac …………………….... (2)

c^{2} = bd …………………...... (3)

It has to be proved that,

(a^{2} + b^{2} + c^{2}) (b^{2} + c^{2} + d^{2}) = (ab + bc – cd)^{2}

**R.H.S.**

= (ab + bc + cd)^{2}

= (ab + ad + cd)^{2} [Using (1)]

= [ab + d (a + c)]^{2}

= (ab)^{2} + 2(ab)d(a + c) + [d(a + c)]^{2}

= a^{2}b^{2} + 2abd (a + c) + d^{2} (a + c)^{2}

= a^{2}b^{2} +2a^{2}bd + 2acbd + d^{2}(a^{2} + 2ac + c^{2})

[Using (1) and (2)]

= a^{2}b^{2} + 2a^{2}c^{2} + 2b^{2}c^{2} + d^{2}a^{2} + 2d^{2}b^{2} + d^{2}c^{2 }[using bc = ad and b^{2} = ac]

= a^{2}b^{2} + a^{2}c^{2} + a^{2}c^{2} + b^{2}c^{2} + b^{2}c^{2} + d^{2}a^{2} + d^{2}b^{2} + d^{2}b^{2} + d^{2}c^{2}

= a^{2}b^{2} + a^{2}c^{2} + (ac)^{2} + b^{2}c^{2} + b^{2}c^{2} + a^{2}d^{2} + (bd)^{2} × (bd)^{2} + c^{2}d^{2}

= a^{2}b^{2} + a^{2}c^{2} + (b^{2})^{2} + b^{2}c^{2} + b^{2}c^{2} + a^{2}d^{2} + (bd)^{2} × (c^{2})^{2} + c^{2}d^{2}

= a^{2}b^{2} + a^{2}c^{2} + b^{2} × b^{2} + b^{2}c^{2} + c^{2}b^{2 }+ a^{2}d^{2} + b^{2}d^{2} + c^{2} × c^{2} + c^{2}d^{2}

= a^{2}b^{2} + a^{2}c^{2} + a^{2}d^{2} + b^{2} × b^{2} + b^{2}c^{2} + b^{2}d^{2} + c^{2}b^{2} + c^{2} × c^{2} + c^{2}d^{2}

[Using (2) and (3) and rearranging terms]

= a^{2}(b^{2} + c^{2} + d^{2}) + b^{2} (b^{2} + c^{2} + d^{2}) + c^{2} (b^{2}+ c^{2} + d^{2})

= (a^{2} + b^{2} + c^{2}) (b^{2} + c^{2} + d^{2}) = L.H.S.

∴ L.H.S. = R.H.S.

∴ (a^{2} + b^{2} + c^{2}) (b^{2} + c^{2} + d^{2}) = (ab + bc – cd)^{2}

**Q26. Insert two numbers between 3 and 81 so that the resulting sequence is G.P.**

**Solution:**

Let G_{1}, G_{2} be three numbers between 3 and 81 such that 3, G_{1}, G_{2}, 81 is a G.P

T_{1} = 3

T_{2} = ar

T_{3} = ar^{2}

T_{4} = ar^{3} = 81

3.r^{3} = 81

r^{3} =

r^{3} = 27

For r = 3, we have

T_{2} = ar = 3.3 = 9

T_{3} = ar^{2} = 3.3^{2} = 27

**Q28. The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio (3 + 2√2) : (3 − 2**√**2).**

**Solution:**

Let the numbers be *a* and *b*.

##### Other Pages of this Chapter: 9. Sequences and Series

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