Q.
In an arithmetic progression containing 99 terms, the sum of all the odd numbered terms is 2550. If the sum of all the 99 terms of the arithmetic progression is k , then 100k is equal to
1723
197
NTA AbhyasNTA Abhyas 2020Sequences and Series
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Answer: 50.49
Solution:
Let the terms of AP be a,a+d,a+2d,…..,a+98d
For sum of all odd numbered terms A=a,D=2d,N=50.
i.e. 250(2a+49(2d))=2550 ⇒2a+98d=102
Sum of all 99 terms is, 299(2a+98d)=299×102=99×51 ⇒k=5049⇒100k=50.49