Question

· 15. In an A.P. ifa p + 1 = 2a q+1,
prove that
a3p + 1 = 2ap+q+1
​

Answer 1

Given:-

• a p + 1 = 2a q+1,

To prove:-

• a3p + 1 = 2ap+q+1

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

Given, a_(p + 1) = 2 a_(q + 1)

or, a + (p + 1 - 1) d = 2 {a + (q + 1 - 1) d}

or, a + pd = 2a + 2qd

or, a + 3pd - 2pd = 2a + 2qd

or, a + 3pd = 2a + 2pd + 2qd

or, a + (3p + 1 - 1) d = 2 {a + (p + q + 1 - 1) d}

or, a_(3p + 1) = 2 a_(p + q + 1)

Hence, proved.

The given sequence is

n, n² - 1, n

The difference of the first two terms,

d₁ = (n² - 1) - n = n² - n - 1 and

the difference between the last two terms,

d₂ = n - (n² - 1) = - n² + n + 1

Since d₁ ≠ d₂, the given sequence is not in A.P.

Hence, proved.

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Hope it helps : )