The sum of the third and seventh term of an A P is 6 and their product is 8 - Maths - Arithmetic Progressions

Question

The sum of the third and seventh term of an A P is 6 and their
product is 8 Find the sum of the first sixteen terms of the A P

Lalit Mehra · Accepted Answer

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

n^th term of A.P., a_n = a + (n – 1) d

∴ a₃ = a + (3 – 1) d = a + 2d

a₇ = a + (7 – 1) d = a + 6d

Given, a₃ + a₇ = 6

∴ (a + 2d) + (a + 6d) = 6

⇒ 2a + 8d = 6

⇒ a + 4d = 3 ...(1)

Given, a₃ × a₇ = 8

∴ (a + 2d) + (a + 6d) = 8

⇒ (3 – 4d + 2d) (3 – 4d + 6d) = 8 [ Using (1) ]

⇒ (3 – 2d) (3 + 2d) = 8

⇒ 9 – 4d² = 8

⇒ 4d² = 1

$\Rightarrow d^{2} = \frac{1}{4} \Rightarrow d = \pm \frac{1}{2} When d = \frac{1}{2},$

a = 3 – 4d $= 3 - 4 \times \frac{1}{2}$ = 3 – 2 = 1

$When d = - \frac{1}{2}, a = 3 - 4 d = 3 - 4 \times (- \frac{1}{2}) = 3 + 2 = 5$

When a = 1 and $d = \frac{1}{2},$

$S_{16} = \frac{16}{2} [2 \times 1 + (16 - 1) \times \frac{1}{2}] = 8 (2 + \frac{15}{2}) = 4 \times 19 = 76 When a = 5 and d = - \frac{1}{2}, S_{16} = \frac{16}{2} [2 \times 5 + (16 - 1) \times (- \frac{1}{2})] = 8 (10 - \frac{15}{2}) = 4 \times 5 = 20$

Thus, the sum of first 16 terms of the A.P. is 76 or 20.

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The sum of the third and seventh term of an A.P. is 6 and their product is 8. Find the sum of the first sixteen terms of the A.P. .