Question

The sum of the first 9 terms of an AP is 81 and that of its first 20 terms is 400. Find the first term and the common difference of the AP.

Answer 1

Hello
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The sum of first 9 terms of an AP = 81

Sum of it's first 20 terms = 400

So,

$s_9 = \frac{9}{2} \times (2a + 8d) \\ \\ = > 81 = \frac{9}{2} \times (2a + 8d) \\ \\ = > \frac{ \cancel{81} \: \: {}^{9} \times 2}{ \cancel 9} = 2a + 8d \\ \\ = > 2a + 8d = 18......(1)$

Again,

$s_{20} = 400 \\ \\ 400 = \frac{20}{2} \times (2a + 19d) \\ \\ = > \frac{400}{10} = 2a + 19d \\ \\ = > 2a + 19d = 40.......(2)$

Now,

Subtracting equations (2) from (1), we get,

2a + 8d = 18

2a + 19d = 40
- - -
_____________

- 11d = - 22

=> 11d = 22

=> d = 22/11

=> d = 2

So ,

Putting d = 2 in equation (1), we get,

2a + 8d = 18

=> 2a + (8 × 2) = 18

=> 2a = 18 - 16

=> a = 2/2

=> a = 1

Therefore,

First term = a = 1

Common difference = d = 2

HOPE IT HELPS

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ArchitectSethRollins

Answer 2

Hey there !!


➡ Given :-

→ S$\tiny 9$ = 81.

→ S$\tiny 20$ = 400.


➡ To find :-

→ First term ( a$\tiny 1$ ) .

→ Common difference (d) .

➡ Solution :-

We have ,

=> S$\tiny 9$ = n/2( 2a ( n - 1 ) d ) .

=> 81 = 9/2 ( 2a + 8d ) .

=> 81 × 2 = 18a + 72d.

=> 162 = 18( a + 4d ) .

=> a + 4d = 9..........(1).

And,

=> S$\tiny 20$ = 20/2( 2a + 19d ).

=> 400 = 10( 2a + 19d ) .

=> 400/10 = 2a + 19d .

=> 2a + 19d = 40........(2).

▶ Multiply equation (1) by 2,

=> 2( a + 4d = 9 ).

=> 2a + 8d = 18..........(3)

▶ On substracting equation (2) and (3), we get

2a + 19d = 40.
2a + 8d = 18.
(-).....(-).......(-)
____________

=> 11d = 22.

$\huge \boxed{ \bf => d = 2. }$


▶ Put the value of ‘d’ in equation (1), we get

=> a + 4 × 2 = 9.

=> a + 8 = 9.

=> a = 9 - 8.

$\huge \boxed{ \bf => a = 1 }$

✔✔ Hence, it is solved ✅✅.

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$\huge \boxed{ \mathbb{THANKS}}$



$\huge \bf{ \#BeBrainly.}$