Porisma

Worked solution

Recover an AP from two consecutive blocks of sums

The first seven terms of an AP sum to 2828, and the next eight terms sum to 6262. List the first three terms of the AP.

CJC · Sequences and Series · Example 20

Answer

52,3,72\frac52,3,\frac72.

Compare your reasoning

Translate the first block

S7=72(2a+6d)=28,7a+21d=28,a+3d=4(1).\begin{aligned}S_7=\frac72(2a+6d)&=28,\\7a+21d&=28,\\a+3d&=4\qquad\text{(1)}.\end{aligned}

Let a be the first term and d the common difference. Retain the source equation (1).

Translate the next eight terms

S15−S7=62,152(2a+14d)−28=62,15a+105d=90,a+7d=6(2).\begin{aligned}S_{15}-S_7&=62,\\\frac{15}{2}(2a+14d)-28&=62,\\15a+105d&=90,\\a+7d&=6\qquad\text{(2)}.\end{aligned}

The second block is terms 8 through 15, so subtract S7 from S15. It is not S8.

Solve and list the requested terms

d=12,a=52,(a,a+d,a+2d)=(52,3,72)d=\frac12,\qquad a=\frac52,\qquad(a,a+d,a+2d)=\left(\frac52,3,\frac72\right)

Subtracting (1) from (2) gives 4d=2. The source gives a=5/2,d=1/2 and the first three terms.

How did this question go?

Common mistakes

A later block of terms is a difference of cumulative sums.