How many terms of the arithmetic progression 45, 39, 33, ........

Q) How many terms of the arithmetic progression 45, 39, 33, …….. must be taken so that their sum is 180? Explain the double answer.

Ans:

We start from the given AP: 45,39, 33, …….

a = 45, d = – 6,

Sum of n terms of AP, S_n = $\frac{n}{2}$ [2 a + (n – 1) d]

180 = $\frac{n}{2}$ [2 (45) + (n – 1)(- 6)]

∴ 360 = n [90 – 6 (n – 1)]

∴ 360 = n (90 – 6 n + 6) = n (96 – 6 n)

∴ 360 = 6 n (16 – n)

∴ 60 = n (16 – n)

∴ 60 = 16 n – n²

∴ n² – 16n + 60 = 0

∴ n² – 10 n – 6 n + 60 = 0 (by mid-term splitting)

∴ n (n – 10) – 6 (n – 10) = 0

∴ (n – 10) (n – 6) = 0

∴ n = 6 and n = 10

Therefore, for sum of first 6 terms is 180 as well as the sum of first 10 terms is 180.

Reason for 2 values:

We get sum of 180 for n = 6 and n = 10 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 = 6, sum of first 6 terms is 180.

This goes on and at n = 8, sum of first 8 terms is 192.

At n = 9, 9th term is – 3. From here, each term becomes a negative value and sum of AP starts to go down.

At n = 9, sum of AP becomes 189 and finally at n = 10, it becomes 180 again.

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