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Question

Find the sum of each arithmetic series
(i) 38+35+32+.....+2
(ii) 6+514+412+....25 terms

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Solution

(i) The given arithmetic series is 38+35+32+....+2 where the first term is a1=38, second term is a2=35 and the nth term is Tn=2.

We find the common difference d by subtracting the first term from the second term as shown below:

d=a2a1=3538=3

We know that the general term of an arithmetic progression with first term a and common difference d is Tn=a+(n1)d, therefore, with a=38,d=3 and Tn=2, we have

Tn=a+(n1)d2=38+(n1)(3)2=383n+32=413n3n=4123n=39n=393n=13

We also know that the sum of an arithmetic series with first term a and common difference d is Sn=n2[2a+(n1)d]

Now to find the sum of series, substitute n=13,a=38 and d=3 in Sn=n2[2a+(n1)d] as follows:

S13=132[(2×38)+(131)(3)]=132[76+(12×3)]=132(7636)=132×40=13×20=260

Hence, the sum of the given arithmetic series is 260.

(ii) The given arithmetic series is 6+514+412+........ where the first term is a1=6, second term is a2=514 and the number of terms n=25.

We find the common difference d by subtracting the first term from the second term as shown below:

d=a2a1=5146=2146=21244=34

We know that the sum of an arithmetic series with first term a and common difference d is Sn=n2[2a+(n1)d]

Now to find the sum of series, substitute n=25,a=6 and d=34 in Sn=n2[2a+(n1)d] as follows:

S25=252[(2×6)+(251)(34)]=252[12+(24×34)]=252[1218]=252×6=25×3=75

Hence, the sum of the given arithmetic series is 75.

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