Let a 1, a 2,.......a 3n be an arithmetic progression with a 1 = 3 and a 2 = 7. If a 1 + a 2 +......+ a 3n = 1830, then what is the smallest positive integer m such that m(a 1 + a 2 +..... + a n ) 1830?

Question

Let a 1, a 2,.......a 3n be an arithmetic progression with a 1 = 3 and a 2 = 7. If a 1 + a 2 +......+ a 3n = 1830, then what is the smallest positive integer m such that m(a 1 + a 2 +..... + a n ) > 1830?

Accepted Answer

9 — Easy Find the common difference from $a_1,a_2$, then use the total $\sum_{1}^{3n}=1830$ to pin down $n$. Compute $\sum_{1}^{n}$, and finally find the smallest $m$ making $m\sum_{1}^{n}$ exceed $1830$. 1 Identify the AP. Common difference $d = a_2-a_1 = 4$, so the $k$-th term is $a_k = 4k-1$. $$\begin{aligned} &a_k = 3 + (k-1)\cdot 4 = 4k-1\ &\Rightarrow\ a_{3n} = 12n - 1 \end{aligned}$$ 2 Use the sum of $3n$ terms to find $n$. $$\begin{aligned} &\frac{3n}{2}\big(a_1 + a_{3n}\big) = 1830\ &\Rig

CAT 2017 Slot 1 — QA Question 34

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