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The sum of the 4th and the 8th terms of an AP is 24 and the sum of its 6th and 10th terms is 44. Then what will be the sum of its first 10 terms?

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Detailed Solution

Given that the sum of the 4th and the 8th terms of an AP is 24 and the sum of its 6th and 10th terms is 44.
Let the first term be a and the common difference be d.
Formula used for nth term of A.P.:

a n = a + n − 1 d

Now, determining 4th term.

∴ a 4 = a + (n − 1) d ⇒ a 4 = a + (4 − 1) d ⇒ a 4 = a + 3 d

Determining 8th term.

∴ a s = a + (n − 1) d ⇒ a 8 = a + (8 − 1) d ⇒ a 8 = a + 7 d

∴ a 4 + a 3 = 24 [given] ⇒ (a + 3 d) + (a + 7 d) = 24 ⇒ 2 a + 10 d = 24 ⇒ a + 5 d = 12 − − − (1)

Also, given that the sum of 6th and 10th term of A.P. is 44.

∴ a 6 + a 10 = 44 ⇒ (a + 5 d) + (a + 9 d) = 44 ⇒ 2 a + 14 d = 44 ⇒ a + 7 d = 22 − − − (2)

Subtracting equation (1) from equation (2):

∴ (a + 7 d) − (a + 5 d) = 22 − 12 ⇒ 7 d − 5 d = 10 ⇒ 2 d = 10 ⇒ d = 5

Therefore, the common difference is 5.
Putting the value of d in equation (1),

∴ a + 5 × 5 = 12 ⇒ a + 25 = 12 ⇒ a = 12 − 25 ⇒ a = − 13

Therefore, the first term is -13.
Formula used for sum of first nth term of A.P.:

S n = n 2 [2 a + (n − 1) d]

Determining the sum of 10 terms:

∴ S n = n 2 [2 a + (n − 1) d] ⇒ S 10 = 102 [2 × (− 13) + (10 − 1) 5] = 5 [− 26 + 45] = 5 [19] = 95

Therefore, the sum of the first 10 terms is 95.
Hence, option (3) is correct.

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