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 = a

_{2}− a_{1}= 1.7 − 0.6 = 1.1n = 100

We know that, the formula for sum of nth term in AP series is,

S

_{n}= n/2[2a +(n-1)d]S= 50/2 [1.2+(99)×1.1]_{12}= 50[1.2+108.9]

= 50[110.1]

= 5505