Question

in an A.P ,if Sn=n(4n+1),find A.P

Answer 1

given ,

Sn =n ( 4n + 1 )

=4n^2 + n

we know ,

Tn = Sn - S(n-1)

=4n^2+n -4 (n-1)^2 - (n-1)

=4 (n^2-n^2+2n-1)+(n-n+1)

=8n - 4 + 1

= 8n -3

hence ,

Tn = 8n -3

T1 =8 (1)-3 =5

T2= 8 (2)-3 =13

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so, AP is 5, 13 , 21 ,...

Answer 2

Answer:

$\huge{\pink{\boxed{\green{\sf{\therefore A.P=5,13,21.....}}}}}$

Step-by-step explanation:

$\huge{\pink{\boxed{\green{\underline{\red{\sf{SOLUTION-}}}}}}}$

$\: \: \: \: \: \: \: {\orange{given}} \\ {\pink{ \boxed{ \green{Sn = n(4n + 1)}}}} \\ \\ {\blue{to \: find}} \\ {\purple{ \boxed{ \red{A.P =? }}}}$

☆ According to given question:

☆ sum of nth term is given.

☆ let n =1,2,3....

$\to Sn = n(4n + 1) \\ \to S1 = 1(4 \times 1 + 1) \\ { \boxed {\to S1 = 5 }}\\ \\ \to S2 = 2(4 \times 2 + 1) \\ \to S2 = 2(9) \\ { \boxed{\to S2 = 18}} \\ \\ \to S3 = 3(4 \times 3 + 1) \\ \to S3 = 3 \times 13 \\ { \boxed{\to S3 = 39 }} \\ \\ { \boxed{ \to s1 = a1 = 5}} \\ \\ \to a2 = S2 - S1 \\ \to a2 = 18 - 5 \\ { \boxed{\to a2 = 13}} \\ \\ \to a3 = S3- S2 \\ \to a3 = 18 - 39 \\ { \boxed{ \to a3 = 21}} \\ \\ { \pink{ \boxed{ \green{\therefore A.P = 5,13,21....}}}}$

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