Question 41

The sum of the $$6^{th}$$ and $$15^{th}$$ elements of an arithmetic progression is equal to the sum of $$7^{th}$$, $$10^{th}$$ and $$12^{th}$$ elements of the same progression. Which element of the series should necessarily be equal to zero?

Solution

Let the A.P. be $$A_1,A_2,A_3,....$$ and so on, with first term = $$a$$ and common difference = $$d$$

Also, $$n^{th}$$ term of an A.P. = $$A_n=a+(n-1)d$$

Acc to ques,

=> $$A_6+A_{15}=A_7+A_{10}+A_{12}$$

=> $$(a+5d)+(a+14d)=(a+6d)+(a+9d)+(a+11d)$$

=> $$a+7d=0$$

Thus, $$A_8=0$$

=> Ans - (B)

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