Consider the arithmetic progressions 3, 7, 11, ... and let An dentoe the sum of the first n terms of this progression. Then the value of 1 25 ∑ A n n = 1 25

Question

Consider the arithmetic progressions 3, 7, 11, ... and let An dentoe the sum of the first n terms of this progression. Then the value of 1 25 &sum; A n n = 1 25

Accepted Answer

455 — Easy Find a closed form for $A_n$ (sum of the first $n$ terms of the AP), then sum $A_n$ from $1$ to $25$ using the standard formulas $\sum n^2=	frac{n(n+1)(2n+1)}{6}$ and $\sum n=	frac{n(n+1)}{2}$. Keep the factor $25$ floating so it cancels neatly at the end. 1 Closed form for $A_n$. First term $a=3$, common difference $d=4$: $$A_n=\frac{n}{2}\big[2\cdot3+(n-1)\cdot4\big]=\frac{n}{2}(4n+2)=2n^2+n.$$ 2 Sum it for $n=1$ to $25$: $$\frac{1}{25}\sum_{n=1}^{25}(2n^2+n)=\frac{1}{25}\Big[2\cdot\fra

CAT 2022 Slot 2 — QA Question 5

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