Question

the sum of first 8 term of an A.P is 100 and sum of first 19 term is 551 . find A.P

Answer 1

Answer:

Step-by-step explanation:

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Let a be first term and d be common difference

A/q

(8/2)[2a+(8-1)d] = 100

⇒2a + 7d =25_____(1)

and (19/2)[2a + (19-1)d] = 551

⇒2a + 18d = 58____(2)

subtracting equation (1) and (2)

11d = 33

⇒d = 3

so a = (25 - 21)/2 = 2

so AP = 2,5,8.11,14.....

Answer 2

Sum of 8th term of an A.P. = 100

$S_{8}$ = 100

Here n = 8

Now ..

$S_{n}$ = $\dfrac{n}{2}$ [2a + (n - 1)d]

$S_{8}$ = $\dfrac{8}{2}$ [2a + (8 - 1)d]

100 = 4 (2a + 7d)

$\dfrac{100}{4}$ = 2a + 7d

25 = 2a + 7d ....(1)

Also;

Sum of first 19 term = 551

Here n = 19

$S_{19}$ = $\dfrac{19}{2}$ [2a + (19 - 1)d]

551 = $\dfrac{19}{2}$ (2a + 18d)

$\dfrac{551\:\times\:2}{19}$ = (2a + 18d)

58 = 2a + 18d .....(2)

Solve (1) & (2) eq. by elimination method ..

+ 58 = + 2a + 18d
+ 24 = + 2a + 07d {Change the signs
______________
+ 33 = 0a + 11d
______________

33 = 11d

11d = 33

$\boxed{d \:=\: 3}$

Put value of d in (1)

25 = 2a + 7(3)

25 = 2a + 21

25 - 21 = 2a

2a = 4

$\boxed{a\:=\:2}$

Now ..

$\textbf{A.P. = a, a + d, a + 2d ...}$

2, 2 + 3, 2 + 3(2) ...

$\textbf{2, 5, 8 ...}$