The question is given from ncert Book of class 10th Chapter no. 5 Ex. 5.3 Q. 1 (iii). In the following question you have to find the sum of the A.P 0.6, 1.7, 2.8 ,…….., to 100 terms. Give the solution briefly.
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Given, 0.6, 1.7, 2.8 ,…, to 100 terms
For this A.P.,
first term, a = 0.6
Common difference, d = a2 − a1 = 1.7 − 0.6 = 1.1
n = 100
We know that, the formula for sum of nth term in AP series is,
Sn = n/2[2a +(n-1)d]
S12 = 50/2 [1.2+(99)×1.1]
= 50[1.2+108.9]
= 50[110.1]
= 5505