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 an Ap.( mene method to sahi lagaya par mera answer correct nahi hai please help me!)​

Answer 1

Answer:

Step-by-step explanation:

Solution :-

Let a be the 1st term and d be the common difference.

We know that,

S(n) = n/2[2a + (n - 1)d]

According to the Question,

1st part,

⇒ 2a + 8d =18 ....... (i)

2nd part,

⇒ 2a + 19d = 40 ...... (ii)

Subtracting eq (i) from (ii), we get

⇒ (2a + 19d) - (2a + 8d) = 40 - 18

⇒ 11d = 22

⇒ d = 2

Putting d's value in Eq (i), we get

⇒ 2a + 16 = 18

⇒ 2a = 18 - 16

⇒ 2a = 2

⇒ a = 2/2

⇒ a = 1

Hence, 1st term is 1 and common difference is 2.

Answer 2

GIVEN :-

$\bullet\ \sf{S_9=81}$

$\sf{\bullet\ S_2_0=400}$

TO FIND :-

The first term and the common difference of the A.P.

SOLUTION :-

$\sf{S_9=\dfrac{n}{2}[2a+(n-1)d]=81 }\\\\\sf{\implies \dfrac{9}{2}(2a+8d) =81}\\\\\sf{\implies \dfrac{1}{2}(2a+8d)=9 }\\\\\sf{\implies 2a+8d=18}\:\:\:\:\:\:\:\: \cdots(i)$

$\rule{150}{2}$

$\sf{S_{20}=\dfrac{n}{2}[2a+(n-1)d]=400 }\\\\\sf{\implies \dfrac{20}{2}(2a+19d)=400 }\\\\\sf{\implies \dfrac{1}{2}(2a+19d)=20}\\\\\sf{\implies 2a+19d=40}\:\:\:\:\:\:\:\: \cdots(ii)$

$\rule{150}{2}$

Subtracting eq.(i) from eq.(ii) :-

$\sf{2a+19d=40}\\\sf{-}\\\sf{2a+8d=18}\\\rule{61}{1}\\\sf{11d=22}\\\\\sf{\implies d=\dfrac{22}{11} }\\\\\boxed{\underline{\boxed{\sf{\implies d=2}}}}$

$\rule{150}{2}$

Putting the value of d in eq.(i) :-

$\sf{2a+8d=18}\\\\\sf{\implies 2a=18-(8\times 2)}\\\\\sf{\implies 2a=2}\\\\\boxed{\boxed{\underline{\sf{\implies a=1}}}}$