7th term of an A.P. is 40. then find the sum of first 13 terms ?

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Salman Ahmad Siddiqui

CSAT Tutor, BPSC DI Tutor (Ex. Drishti IAS Senior Content Writer)

7th term =40=a + 6d sum of 13 term =(13/2)(2a+12d)=13(a+6d)=13*40=520
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JU-IITKGP & 10+ years experienced for boards, JEE,NEET,OLYMPIAD & 5-12

a+6d= 40 S13= 13/2(2a+12d) = 13(a+6d) = 13*40 = 520 let series a,a+d,a+2d,............a+nd 7th term=a+6*d=7 sum of 13 terms=13/2*(a+a+12*d)=13*(a+6*d)=520
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Maths lover

let series a,a+d,a+2d,............a+nd 7th term=a+6*d=7 sum of 13 terms=13/2*(a+a+12*d)=13*(a+6*d)=520
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Home Tutor

520
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JU-IITKGP & 10+ years experienced for boards, JEE,NEET,OLYMPIAD & 5-12

(13/2) 2(a+6d) = (13) (40) =520
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For the given A,P. a+6d=40, Then sum of 13 terms of this A.P. is 13/2( 2a+12d) =13(a+6d) = 13(40) = 520.
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Aspiring Mathematician

Given, 7th term of an AP = a+ 6d= 40 where a and d are first term and common difference in the AP. Sum of first 13 terms = (13/2)*(2a+12d) = 13*(a+6d) = 13*40 (since a+6d = 40) = 520
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I am a Master

First term =a Common differnce=d 40=a+(7-1)d ...... (1) 40=a+6d For sum of first 13 terms Sn=n/2 (2a+(n-1)d) Sn=13/2(2a+(13-1)d) Sn=13/2 (2a+(12)d) By taking 2 common we get Sn=13 (a+6d) From 1 Sn =13×40=520
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Mathematics is not about numbers, equation, computation, or algorithms it is about UNDERISTANDING.

sum of first 13 terms is 520
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VIJAY RAY

Ans. a7=40 expanding with formula : An = A + (N-1)D a+6d=40 ......... eq.- 1 Now, S13 = ? expanding by formula : Sn = N/2 (2A+ (N-1)D) =13/2 (2a+12d) =13/2*2 (a+6d) ...............(taking 2 common) =13 (a+6d) =13*40 ................(from eq. 1) =520
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