If the square of the 7 th term of an arithmetic progression with positive common difference equals the product of the 3 rd and 17 th terms, then the ratio of the first term to the common difference is:

Question

Accepted Answer

2 : 3 — Easy Write each term as first term $a$ plus a multiple of the common difference $d$. Translate the given condition into an equation, expand, and solve for the ratio $a:d$. 1 Set up the terms. The $k$-th term is $a+(k-1)d$. $$\begin{aligned} &a_7 = a+6d,\quad a_3 = a+2d,\quad a_{17} = a+16d \end{aligned}$$ 2 Apply the condition $a_7^{\,2} = a_3\cdot a_{17}$ and expand. $$\begin{aligned} &(a+6d)^2 = (a+2d)(a+16d)\ &\Rightarrow\ a^2 + 12ad + 36d^2 = a^2 + 18ad + 32d^2\ &\Rightarrow\ 4d^2 = 6ad \e

CAT 2017 Slot 1 — QA Question 31

Answer & solution

Related Arithmetic Progression questions