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In arithmetic progression, the common difference is 4/3 times of the first term and the sum of the first seventeen terms is 595/12. Then, what is the fifth term?

In arithmetic progression, the common difference is 4/3 times of the first term and the sum of the first seventeen terms is 595/12. Then, what is the fifth term? Correct Answer 19/12

Given:

d = 4a/3

S17 = 595/12

Formula used:

Sn = n/2 ×

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

Calculation:

S17 = 17/2 ×

⇒ 595/12 = 17/2 ×

⇒ 35/6 = 2a + 64a/3

⇒ 35/6 = 70a/3

⇒ a = 1/4

∴ d = 4a/3 = 1/3

So, T5 = a + (n – 1) d

⇒ T5 = ¼ + (5 – 1) × (1/3)

⇒ T5 = ¼ + 4/3

∴ T5 = 19/12

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