Question

only genius can ans. this .....no spammimg....

Answer 1

Hey mate,

Here is ur answer,

Sum of 9 terms => 81.

$sn = \frac{n}{2} (2a + (n - 1)d)$

$= > \: \frac{9}{2} (2a + (9 - 1)d) = 81$
$= > \: a + 4d = \: 9...............(1)$


Sum of 20 terms => 400.

$= > \: \frac{20}{2} (2a + (20 - 1)d) = 400$

$= > \: 2a + 19d = 40..........(2)$

Multiply the equation (1) by 2.

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

Subtracting eq (3) from (2).

=> 2a+19d-2a-8d =40-18.

=> 11d =22.

=> d=2.

Substitute the value of d in eq (1).

=> a+4d = 9.

=> a+4 (2) =9.

=> a =9-8 =1.

Sum of first 15 terms

$s15 = \: \frac{15}{2} (2(1) + 14(2))$

$= > \: \frac{15}{2} (2 + 28)$

=> 15×30/2.

=> 15×15

=> 225.

So, The first term is 1.


Common difference => 2.

Sum of first 15 terms => 225.


:-(Hope it helps u. 《《 VIDYA》》

Answer 2

Hope it helps.........