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Class 12 Tuition > Learn Class XI-XII Tuition (PUC) > 7th term of an A.P. is 40. then find the sum of first...
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7th term of an A.P. is 40. then find the sum of first 13 terms ?

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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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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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520
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Math magacian

a+6d= 40 ....(1) S13 = 13/2 = 13/2*2 ........(2) = 13 *40. From eq (1) ' = 520
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given 7th term = 40 i.e. a + 6d = 40 sum to 13 term = (13/2) (2a + 12 d) = (13/2) 2(a+6d) = (13) (40) =520
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Sum of 13 terms =middle term×no. of terms=7th term×13=40×13=520
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Problem Cracker

tn=a+(n-1)*d where tn is the nth term, here it is 40, a is the first term, n is the term no, here n=7 and d is the common interval. so, a+(7-1)*d=40 a+6d=40----------------------------(i) now the sum of nth term is Sn= (n/2)*{2a+(n-1)*d} = (13/2)*{2a+(13-1)*d} =6.5*(2a+12d) ...
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tn=a+(n-1)*d where tn is the nth term, here it is 40, a is the first term, n is the term no, here n=7 and d is the common interval. so, a+(7-1)*d=40 a+6d=40----------------------------(i) now the sum of nth term is Sn= (n/2)*{2a+(n-1)*d} = (13/2)*{2a+(13-1)*d} =6.5*(2a+12d) =6.5*2(a+6d)--------------------------(ii) now putting the value of equation (i) in equation (ii) we get, Sn= 6.5*2*40 =520 read less
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a+6d= 40 S13= 13/2(2a+12d) = 13(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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