Question

The sum of the 3rd and 7th term.Of an



a.P is 6 and their product is 8 find the sum of first 15 terms of



a.P

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Given :-

→ a₃ + a₇ = 6.

→ a₃ × a₇ = 8 .

To find :-

→ S₁₅ .

Solution :-

We have,

→ a₃ + a₇ = 6.

⇒ a + 2d + a + 6d = 6 .

⇒ 2a + 8d = 6 .

⇒ 2( a + 4d ) = 6 .

⇒ a + 4d = 6/2 .

⇒ a + 4d = 3

∵a = 3 - 4d ............(1) .

And,

→ a₃ × a₇ = 8 .

⇒ ( a + 2d ) × ( a + 6d ) = 8 .

⇒ ( 3 - 4d + 2d ) ( 3 - 4d + 6d ) = 8 .

⇒ ( 3 - 2d )( 3 + 2d ) = 8 .

⇒ 3² - (2d)² = 8 .

⇒ 9 - 4d² = 8.

⇒ 4d² = 9 - 8 .

⇒ 4d² = 1 .

⇒ d² = 1/4 .

⇒ d = √(1/4) .

∴ d = 1/2 .

Putting the value of d in equation (1), we get ,

⇒ a = 3 - 4 × 1/2 .

⇒ a = 3 - 2 .

∴ a = 1 .

Thus, sum of 15th term is given by ,

$\because \sf S_{15} = \frac{n}{2} \{2a + (n - 1)d \}. \\ \\ \sf = \frac{15}{2} \{ 2 \times 1 + (15 - 1) \frac{1}{2} \}. \\ \\ \sf = 7.5 \{2 + \frac{14}{2} . \} \\ \\ \sf= 7.5 \bigg( 2 +7 \bigg). \\ \\ \sf = 7 5 \times 9. \\ \\ \huge \pink{ \boxed{ \tt\therefore S_{15} = 67.5.}}$

Hence, sum of first 15 terms is 67.5.