Q) How many terms of the arithmetic progression 24, 21, 18, …….. must be taken so that their sum is 78? Explain the double answer.  

Ans:

We start from the given AP: 24, 21, 18, …….

a = 24, d = – 3, S_n = 78 (given)

Sum of n terms of AP, S_n = [2 a + (n – 1) d]

∴ 78 = [2 (24) + (n – 1)(- 3)]

∴ 156 = n [48 – 3 (n – 1)]

∴ 156 = n (48 – 3 n + 3)

∴ 156 = n (51 – 3 n)

∴ 156 = 3 n (17 – n)

∴ 52 = n (17 – n)

∴ 52 = 17 n – n²

∴ n² – 17n + 52 = 0

∴ n² – 13 n – 4 n + 52 = 0         (by mid-term splitting)

∴ n (n – 13) – 4 (n – 13) = 0

∴ (n – 13)(n – 4) = 0

∴ n = 4 and n = 13

Therefore, for sum of first 4 terms is 78 as well as the sum of first 13 terms is 78.

Reason for 2 values:

We get sum of 78 for n = 4 and n = 13 because the AP is in decreasing order

Explanation: In the given AP, the value of each term is lower than the previous one. 

At n = 4, sum of first 4 terms is 78 (= 24 + 21 + 18 + 15)

This goes on and at n = 8, sum of 8 terms is 108.

9th term is 0 and 10th term is – 3 and from here, each term becomes a negative value. 

Hence, at n = 10, sum of AP starts to go down, and finally at n = 13, it becomes 78 again. 

Therefore, sum of first 4 terms and sum of 13 terms are 78.

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