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In arithmetic progression, the first term is 32 less than the fifth terms and the common difference 4/21 times of the first term of A.P., then, find the sum of the first twelve terms?
1. 1050
2. 1064
3. 1032
4. 1024

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Correct Answer - Option 3 : 1032

Given:

The first term, a = T5 – 32

The common difference, d = 4a/21

Formula used:

Tn = a + (n – 1) d

Sn = n/2 × [2a + (n – 1) d]

Where, a = first term, n = number of terms and d = common difference

Calculation:

T5 = a + (5 – 1) d

⇒ T5 = a + 4d

According to question-

⇒ a + 4d – 32 = a

⇒ 4d = 32

⇒ d = 32/4

⇒ d = 8

Now, d = 4a/21

⇒ 8 = 4a/21

⇒ a = 42

Therefore, S12 = 12/2 × [2(42) + (12 – 1) 8]

⇒ S12 = 6 × [84 + 88]

⇒ S12 = 6 × 172

∴ S12 = 1032

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