Question

please answer this question ​

Answer 1

Step-by-step explanation:

S9 = 81

S20 = 400

first term a=?

common difference d =?

sum of first 9 terms

S9 = 9/2[2a+(9-1)d]

81 = 9/2[2a+8d]

81/9 = (a+4d)

a+4d = 9....................1

sum of first 20 terms

S20 = 20/2[2a+(20-1)d]

400 = 10[ 2a+19d]

400/10 = 2a+19d

......................2

by elimination method

a+4d = 9. ×2

2a + 8d= 18

2a+19d = 40

-. - -

__________

0 - 11d = -22

d = -22/-11

d = 2

put value of d in equation 1

a+4d= 9

a +4×2=9

a+8 = 9

a = 9-8

a = 1

Answer 2

Step-by-step explanation:

Question:-

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

Solution:-

As, we know that :-

The formula for finding the Sum of n terms is:-

$\boxed{ \bf \longrightarrow S_{n} = \frac{n}{2} \big[ 2a + (n - 1)d\big]}$

Sum of 9 terms = 81

$\dashrightarrow \sf S_{9} = 81$

$\dashrightarrow \sf \dfrac{9}{2} \big[ 2a + (9 - 1)d\big]= 81$

$\dashrightarrow \sf \dfrac{9}{2} \big[ 2a + 8d\big]= 81$

$\dashrightarrow \sf 2a + 8d= 81\times \dfrac{2}{9}$

$\dashrightarrow \sf 2a + 8d= 18......(i)$

Sum of 20 terms = 400

$\dashrightarrow \sf S_{20} = 400$

$\dashrightarrow \sf \dfrac{20}{2} \big[ 2a + (20 - 1)d\big]= 400$

$\dashrightarrow \sf 10( 2a + 19d) = 400$

$\dashrightarrow \sf 2a + 19d = \dfrac{400}{10}$

$\dashrightarrow \sf 2a + 19d=40......(ii)$

$\sf Equation \: (ii) - (i)$

$\dashrightarrow \sf 2a + 19d - (2a + 8d) = 40 - 18$

$\dashrightarrow \sf 2a + 19d - 2a - 8d = 22$

$\dashrightarrow \sf 11d = 22$

$\dashrightarrow \sf d = \dfrac{22}{11}$

$\dashrightarrow \boxed{\sf d = 2}$

$\sf Substituting \: d = 2 \: in\: equation\: (i)$

$\dashrightarrow \sf 2a + 8d= 18$

$\dashrightarrow \sf 2a + 8(2) = 18$

$\dashrightarrow \sf 2a = 18 - 16$

$\dashrightarrow \sf 2a = 2$

$\dashrightarrow \sf a = \dfrac{2}{2}$

$\dashrightarrow \boxed{\sf a = 1}$

$\leadsto \underline{ \boxed{ \pink{ \therefore\textsf{ \textbf{First \: term \: = \: 1}}}}}$

$\leadsto \underline{ \boxed{ \purple{ \therefore\textsf{ \textbf{Common \: difference \: = \: 2}}}}}$