Question

An AP has 21 terms. The sum of the 10th, 11th, and 12th term is 129 and the sum of the last 3 terms is 237. Find the AP

PLEASE SOME HELP ME WITH A SOLVED ANSWER OF THIS QUESTION ASAP!!

Answer 1

For n consecutive nos. of an AP, if n is odd, their sum will be the product of n and the middle term.

$T_{10} + T_{11} + T_{12} = 129 \\ = 3 * T_{11} = 129 \\ \\ T_{11} = \frac{129}{3} = 43$

Sum of last 3 terms,

$T_{19} + T_{20} + T_{21} = 237 \\ = 3 * T_{20} = 237 \\ \\ T_{20} = \frac{237}{3} = 79$

$T_{20} - T_{11} = 79 - 43 \\ = (20 - 11)d = 36 \\ = 9d = 36 \\ \\ d = \frac{36}{9} = 4$

4 is the common difference.

Let's find first term.

$T_{11} - T_{1} = (11 - 1)d \\ = 43 - T_{1} = 10 * 4 \\ = 43 - T_{1} = 40 \\ \\ T_{1} = 43 - 40 = 3$

3 is the first term.

∴ The AP is 3, 7, 11, 15,......, 75, 79, 83.

Hope this may be helpful.

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