Question

Find the sum of 25th term and if a 3 is equal to 7 and seventh term is equal to 2 more than thrice of its third term

Answer 1

Given.

Third term, a₃ = 7

Seventh term, a₇ = 3a₃ + 2

= 3 × 7 + 2 = 23

To find.

Sum of 25 terms, S₂₅

Solution.

Let a and d be the first term and common difference of the AP.

a₃ = 7

⇒ a + (3 – 1)d = 7

⇒ a + 2d = 7 …(1)

Also, a₇ = 23

⇒ a + (7 – 1)d = 23

⇒ a + 6d = 23 …(2)

Subtracting (1) from (2),

4d = 16

⇒ d = 4

Put d = 16 in (1)

a + (2 × 4) = 7

⇒ a + 8 = 7

⇒ a = –1

Now, S₂₅

$= \frac{25}{2} (2a + 25d - d)$

$= \frac{25}{2}(2a + 24d)$

$= \frac{25}{2}( - 1 \times 2 + 24 \times 4)$

$= \frac{25}{2}( - 2 + 96)$

$= \frac{25}{2}(94)$

$= 25 \times 47$

$= 1175$

∴ Sum of first 25 terms is 1175.

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