Find the sum of the arithmetic sequence 140, 134, 128…, if there are 33 terms

limfne2c
2022-10-17
Answered

Find the sum of the arithmetic sequence 140, 134, 128…, if there are 33 terms

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Martha Dickson

Answered 2022-10-18
Author has **20** answers

We first work out the 33th term, the general formula for a sequence like this being ${a}_{n}={a}_{0}+n\times d$, d being the difference.

Every term is the previous plus or minus the differnce d=−6. So to get to the 33th term we take 32 of these steps (remember we call the first term $a}_{0$ and the 33th is thus $a}_{32$ -- remember this!!):

${a}_{32}=140+32\times (-6)=-52$

The sum is $33\times$ the average, or

$\sum =33\times \frac{140+(-52)}{2}=33\times \frac{88}{2}=1452$

Every term is the previous plus or minus the differnce d=−6. So to get to the 33th term we take 32 of these steps (remember we call the first term $a}_{0$ and the 33th is thus $a}_{32$ -- remember this!!):

${a}_{32}=140+32\times (-6)=-52$

The sum is $33\times$ the average, or

$\sum =33\times \frac{140+(-52)}{2}=33\times \frac{88}{2}=1452$

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