Question

.If the sum of the first n terms of an AP is 3n2 + 1 and its common difference is 6, then

find the first term​

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{First\:term=4}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies Sum \: of \: n \: terms \: of \: an \: A.P(s_{n} )= 3 {n}^{2} + 1 \\ \\ \red{\underline \bold{To \: Find:}} \\ \tt: \implies First \: term(a_{1}) =?$

• According to given question :

$\tt \circ \: s_{n} = 3 {n}^{2} + 1 \\ \\ \tt\because \: n = 1,2,3..... \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies s_{n} = {3n}^{2} + 1 \\ \\ \tt: \implies s_{1} = 3 \times {1}^{2} + 1 \\ \\ \tt: \implies s_{1} = 3 + 1 \\ \\ \green{\tt: \implies s_{1} = 4 = a_{1} } \\ \\ \green{\tt \therefore First \: term \: of \: A.P \: is \: 4}\\\\ \blue{ \bold{Some \: related \: formula}} \\ \orange{\tt \circ \: a_{n} = a + (n - 1)d} \\ \\ \orange{\tt \circ \:s_{n} = \frac{n}{2}(a + l)}$

Answer 2

GIVEN :-

• The sum of the first n terms of an AP is 3n² + 1.

• Common difference is 6.

TO FIND :-

The first term of the A.P.

SOLUTION :-

We know,

$\sf{S_n=\dfrac{n}{2}[2a+(n-1)d] }$

$\sf{\implies 3n^2+1=\dfrac{n}{2}[2a+(n-1)6] }\\\\\sf{\implies 6n^2+2=n(2a+6n-6)}\\\\\sf{\implies 6n^2+2=2an+6n^2-6n}\\\\\sf{\implies 2an-6n-2=0}\\\\\sf{\implies 2an=6n+2}\\\\\sf{\implies 2an=2(3n+1)}\\\\\sf{\implies an=3n+1}\\\\\sf{\implies a=\dfrac{3n+1}{n} }\\\\\sf{\implies a_1=\dfrac{3n+1}{n}}\\\\\sf{\implies a_1=\dfrac{3(1)+1}{1} }\\\\\boxed{\sf{\implies a_1=4}}$

Therefore, the first term of the A.P. is 4.