Math, asked by niteshshaw723, 9 months ago

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Answered by shashiawasthi069
2

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

Answered by MaIeficent
9

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}}}}}

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