Question

the sum of frist n terms of 3 ap are s1 S2 S3 res. the first term of each ap is 1 and common difference are 1,2,3 res. prove that S1 + S3= 2S2

Answer 1

There you go pal,

Nice question.

Answer in attachment

Answer 2

$\underline{\underline{\Large{\mathfrak{Solution : }}}}$




$\underline{\textsf{For 1st A.P.,}} \\ \\ \sf \implies First \: term(a_1) \: = \: 1 \\ \\ \sf \implies Common \: different (d_1 ) \: = \: 1 \\ \\ \sf \implies Number \: of \: terms \: = \: n \\ \\ \textsf{Now, } \\ \\ \sf \implies S_1 \: = \: \dfrac{n}{2}\{2a_1 \: + \: ( n \: - \: 1 )d_1 \} \\ \\ \sf \implies S_1 \: = \: \dfrac{n}{2}\{ 2 \: \times \: 1 \: + \: ( n \: - \: 1)1 \} \\ \\ \sf \implies S_1 \: = \: \dfrac{n}{2}(2 \: + \: n \: - \: 1 )\\ \\ \sf \: \: \therefore \: \: S_1 \: = \: \dfrac{n}{2}(n \: + \: 1 ) \qquad...(1)$




$\underline{\textsf{For 2nd A.P.,}} \\ \\ \sf \implies First \: term(a_2) \: = \: 1 \\ \\ \sf \implies Common \: different (d_2 ) \: = \: 2 \\ \\ \sf \implies Number \: of \: terms \: = \: n \\ \\ \textsf{Now, } \\ \\ \sf \implies S_2 \: = \: \dfrac{n}{2}\{2a_2 \: + \: ( n \: - \: 1 )d_2 \} \\ \\ \sf \implies S_2 \: = \: \dfrac{n}{2}\{ 2 \: \times \: 1 \: + \: ( n \: - \: 1)2\} \\ \\ \sf \implies S_2 \: = \: \dfrac{n}{2}( \cancel{2 }\: + \: 2n \: - \: \cancel{ 2})\\ \\ \sf \implies S_2 \: = \: \dfrac{n}{ \cancel{2}} \: \times \: \cancel{2}n \\ \\ \sf \: \: \therefore \: \: S_2 \: = \: {n}^{2} \qquad...(2)$




$\underline{\textsf{For 3rd A.P.,}} \\ \\ \sf \implies First \: term(a_3) \: = \: 1 \\ \\ \sf \implies Common \: different (d_3 ) \: = \: 3 \\ \\ \sf \implies Number \: of \: terms \: = \: n \\ \\ \textsf{Now, } \\ \\ \sf \implies S_3 \: = \: \dfrac{n}{2}\{2a_3 \: + \: ( n \: - \: 1 )d_3\} \\ \\ \sf \implies S_3 \: = \: \dfrac{n}{2}\{ 2 \: \times \: 1 \: + \: ( n \: - \: 1)3 \} \\ \\ \sf \implies S_3 \: = \: \dfrac{n}{2}(2 \: + \: 3n \: - \: 3)\\ \\ \sf \: \: \therefore \: \: S_3 \: = \: \dfrac{n}{2}(3n \: - \: 1 ) \qquad...(3)$




$\underline{\textsf{Now , add (1) and (3) , }} \\ \\ \sf \implies S_1 \: + \: S_3 \: = \: \dfrac{n}{2}(n \: + \: 1) \: + \: \dfrac{n}{2}(3n \: - \: 1 ) \\ \\ \sf \implies S_1 \: + \: S_3 \: = \: \dfrac{n}{2}( n \: + \: \cancel{1} \: + \: 3n \: - \: \cancel{ 1 }) \\ \\ \sf \implies S_1 \: + \: S_3 \:= \: \dfrac{n}{2}( 4n) \\ \\ \sf \implies S_1 \: + \: S_3 \: = \: n \: \times \: 2n \\ \\ \sf \: \: \therefore \: \: S_1 \: + \: S_3 \: = \: 2n^2 \qquad...(4)$




$\underline{\textsf{Now , divide (4) by (2), }} \\ \\ \sf \implies \dfrac{S_1 \: + \: S_3}{S_2} \: = \: \dfrac{2 \: \cancel{n^2}}{\cancel{n^2}} \\ \\ \sf \implies \dfrac{S_1 \: + \: S_3}{S_2} \: = \: 2 \\ \\ \sf \: \: \therefore \: \: S_1 \: + \: S_3 \: = \: 2S_2 \\ \\ \\ \underline{\underline{\mathsf{\Large{Proved !! }}}}$