Chapter 6 Sequence And Series Solutions SEBA

EXERCISE 6.2

1. Choose the arithmetic progressions from the following following sequences:

(i) 4, 8, 12, 16, 20, ...

(ii) $\frac{5}{2}$, $\frac{3}{2}$, $\frac{1}{2}$, $-\frac{1}{2}$, $-\frac{3}{2}$, ... ...

(iii) $1^2$, $2^2$, $3^2$, $4^2$, $5^2$, ... ...

(iv) -2$\frac{3}{5}$, $-\frac{21}{10}$, $-\frac{8}{5}$, $-\frac{11}{10}$, $-\frac{3}{5}$, ... ...

(v) 1 - 2x, 1 + x, 1, 1 - x, -1 - 2x, ... ....

(vi) 0.8, 2.8, 3.8, 5.8, 4.8, ... ...

(vii) 0, 3, 8, 15, 24, ... ...

(viii) 4, 7, 10, 13, 16, ... ...

Solution: (i) We have, 8 - 4 = 4

12 - 8 = 4

16 - 12 = 4

20 - 16 = 4

Hence, the sequence is in AP.

(ii) We have,

Hence, the sequence is in AP.

(iii) We have,

Hence, the sequence is not in AP.

(iv) We have,

Hence, the sequence is not in AP.

(v) We have, (1+x) - (1-2x) = 3x

1 - (1+x) = -x

(1-x) -1 = -x

(-1-2x) - (1-x) = -1 -2x -1 + x = -2 - x

Hence, the sequence is not in AP.

(vi) We have, 2.8 -0.8 = 2

3.8 - 2.8 = 1

5.8 -3.8 = 2

4.8 - 5.8 = -1

Hence, the sequence is not in AP.

(vii) We have, 3 - 0 = 3,

8 - 3 = 5,

15 - 8 = 7,

24 - 15 = 9,

Hence, the sequence is not in AP.

(viii) We have, 7 - 4 = 3,

10 -7 = 3,

13-10 = 3,

16-13 = 3,

Hence, the sequence is in AP.

2. Find the terms mentioned in the brackets of the following sequences -

(a) 9, 7, 5, 3, ... ... ... (15th term)

Solution: We have, the first term a = 9,

common difference, d = 7 - 9 = -2

Then, $t_{15 }$ = a + (15 -1)d

= 9 + 14 x (-2)

= 9 - 28

= -21

(b) 2a-b, 3a, 4a+b, ... ... ... (20th term)

Solution: We have, the first term, $t_{1 }$ = 2a - b,

common difference, d = 3a - (2a-b) = a+b

Then, $t_{20 }$ = $t_{1 }$ + (20 -1)d

= 2a - b + 19 x (a+b)

= 2a - b + 19a+ 19b

= 21a + 18b

(c) $\frac{1}{5}$, $\frac{4}{5}$, $\frac{7}{5}$, $\frac{10}{5}$, ... ... (11th term)

Solution: We have, the first term, a= $\frac{1}{5}$,

common difference, d = $\frac{4}{5}$ - $\frac{1}{5}$ = $\frac{3}{5}$

Then, $t_{11 }$ = a + (10 -1)d

= $\frac{1}{5}$ + 9 x $\frac{3}{5}$

= $\frac{28}{5}$

(d) 10, - 2, -14, ... ... (9th term)

Solution: We have, the first term a = 10,

common difference, d = -2 - 10 = -12

Then, $t_{9}$ = a + (9 -1)d

= 10 + 8 x (-12)

= 10 - 96

= - 86

(e) x - 2, 4x, 7x + 2, ... ... (nth term)

Solution: We have, the first term a = x - 2,

common difference, d = 4x - (x - 2) = 3x + 2

Then, $t_{n }$ = a + (n -1)d

= (x - 2) + (n-1)(3x + 2)

= x - 2 + n(3x + 2) - 3x - 2

=-2x - 4 + n(3x + 2)

(f) 0.02, 0.10, 0.18, .26, .34, ... ... (17th term)

Solution: We have, the first term, a = 0.02

common difference, d = 0.10 - 0.02 = 0.08

Then, $t_{17 }$ = a + (17 -1)d

= 0.02 +16 x 0.08

= 0.02 + 1.28

= 1.3

3. A few pairs of numbers are given below. Find the arithmetic means as mentioned in the brackets with the pairs.

(i) 8 and -7 (4 AMs)

Solution:

(ii) 14 and -10 (5 AMs)
Solution:

(iii) 3 and 21 (2 AMs)

Solution:

(iv) $\frac{3}{7}$ and $\frac{23}{7}$ (9 AMs)
Solution:

(v) -6 and 26 (7 AMs)

Solution:

(vi) 1.2 and 8.4 (5 AMs)

(vii) x and 9x + 4 (3 AMs)

Solution:

4. If m+2, 4m - 6 and 3m - 2 are three consecutive terms of an AP, then find the value of m.

5. I an AP, the fourth term is 10 times the first term. Then show that its 6th term is 4 times the second term.

6. Find the term mentioned in the bracket in each case below and also find the sum of the sequence upto that term -

(i) -26, -23, -20, ... ... (21st term)

(ii) 5.7, 6.9, 8.1, ... ... (31st term)

(iii) 5n-1, 7n, 9n+1, 11n+2, ... ... (27th term)

(iv) 1, 3, 5, 7, ... ... (100th term)

(v) -3$\frac{1}{3}$, -3, -2$\frac{2}{3}$, ... ... (11th term)

7. If a, b, c are in AP then show that the following terms are also in AP.

(i) $\frac{1}{bc}$, $\frac{1}{ca}$, $\frac{1}{ab}$

(ii) a$(\frac{1}{b}+\frac{1}{c})$, b$(\frac{1}{c}+\frac{1}{a})$, c$(\frac{1}{a}+\frac{1}{b})$

(iii) $a^2(b + c)$, $b^2(c+a)$, $c^2(a+b)$

(iv) ab(a + b), bc(b + c), ca(c + a)

Solution:

(v) $\frac{1}{ \sqrt b + \sqrt c}$, $\frac{1}{ \sqrt c + \sqrt a}$, $\frac{1}{ \sqrt a + \sqrt b}$

Solution:

(vi) $(b + c)^2-a^2$, $(c+a)^2-b^2$, $(a+b)^2-c^2$

8. (i) If $\frac{1}{a}$, $\frac{1}{b}$, $\frac{1}{c}$ are in AP, then prove that $\frac{b + c}{a}$, $\frac{c+a}{b}$, $\frac{a+b}{c}$ are also in AP.

(ii) If b+c, c+a, a+b are in AP, then show that a$(\frac{1}{b}+ \frac{1}{c})$, b$(\frac{1}{c}+ \frac{1}{a})$, c$(\frac{1}{a}+ \frac{1}{b})$ are also in AP.

9. If a, b, c are in AP then find the arithmetic means of the following pairs:

(i) (b + c) and (a + b)

(ii) $\frac{1}{bc}$ and $\frac{1}{ab}$

(iii) $\frac{a(b + c)}{bc}$ and $\frac{c(a + b)}{ab}$

(iv) $a^2-(b + c)^2$ and $c^2-(a+b)^2$

10. In an AP, first term is 24 and common difference is - 4, find the number of terms of the AP to be added to get the sum 72.

11. An AP is -9, -6, -3, ... ... How many terms of the AP are to be added to get the sum 66?

12. The sum and product of three consecutive terms of an AP are 27 and 504 respectively. Find the terms.

13. An AP consist of 21 terms. If the sum of three middle term is 129 and the sum of the last three terms is 237, then find the AP.

14. $85^2-83^2+84^2-82^2+83^2-81^2+82^2-80^2+ ...$ ... is a series. Find the sum of the series upto 30 terms.

15. In an AP, if m times the m th term is equal to n times the n th term, then find its (m+n)th term.

16. If the sum of 6 consecutive terms of an AP is 345 and $t_6-t_1=55$; then find the terms.

17. The sum of 4 consecutive terms o an AP is 32 and the sum of their squares is 276; find the terms.

18. In an AP, $S_n = n^2P$ and $S_m = m^2P$, (m $\neq$ n), then show that $S_P = p^3$.

19. The measures of the three angles of a triangle are in AP. If the measure of the biggest angles is twice the measure of the smallest angle, then find the angles of the triangle.

20. The ratio of the sums to n terms of two arithmetic progressions is (7n+1):(4n+27). Find the ratio of their 11th terms.

21. In an AP, prove that $t_p+t_{p+2q}=2.t_{p+q}$.

22. In an AP, if m th term is 'n' and n th term is 'm' then find the pth and the (m+n)th terms of the AP.

23. If $\alpha$ and $\beta$ are the p th and q th terms of an AP, then show that the sum of first (p+q) terms of the AP is $\frac{1}{2}$(p+q)($\alpha$+$\beta$+$\frac{\alpha - \beta}{p - q}$).

24. If first, second and n th terms of an AP are a,b and l respectively, then show that $n=\frac{2a-b-l}{a-b}$ and $S_n =$$\frac{(a+l)(2a - b - l}{2(a-b)}$.

25. If in an AP, $T_q = 2T_p$ and $T_r = 2T_q$ then prove that 2p - 3q + r = 0 (Here $T_p$, $T_q$, $T_r$ represent p th, q th, r th terms of the AP respectively).

26. The first term and common difference of an AP are respectively a and d. In another AP, the first term and the common difference are respectively a' and d'. If $S_m=S_m'$ then prove that m = 1 + $\frac{2(a-a')}{d'-d}$.

27. In three arithmetic progressions their first terms are same and their common differences are 2, 4 and 6 respectively. If sum of n terms of each of the series are respectively $S_1$, $S_2$, $S_3$ then show that 2$S_2$ = $S_1$ + $S_3$.

28. (i) Find the sum of natural numbers lying between 250 and 1000, which are divisible by 3.

(ii) Find the sum of the integers lying between 1 and 100 and which are divisible by 2 or 5.

29. If the sum of the n even natural numbers is $S_1$ and the sum of the first n odd natural numbers is $S_2$, then prove that, n.$S_1$=(n+1)$S_2$.

30. A man has borrowed Rs. 4050 from another man on condition of repaying on monthly instalments. He has paid Rs. 200 as first instalment. After that in each of the remaining instalments he has paid Rs. 25 more than the preceeding instalment. Find the number of instalments required to repay the entire amount.