Question

27. The 4th term of an AP is zero. Prove that its 25th term is triple its 11th term.
(CBSE 2005)​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

SOLUTION :-

Given that,

4th term = 0

Let first term be a and common difference be d.

We know that,

$\bf a_n=a+(n-1)d$

$\bullet\sf \ a_4 = 0$

 $\longrightarrow \sf a+(4-1)d= 0 \\\\\longrightarrow a+3d= 0 \\\\\longrightarrow\bf a = -3d$

We have to prove that,

25th term = 3 × 11th term

$\bullet\sf \ L.H.S = a_{25}$

            $=\sf a+(25-1)d$

            $=\sf a+24d$

            $=\sf -3d+24d$

            $=\bf 21d$

$\bullet\sf \ R.H.S = 3\times a_{11}$

            $=\sf 3\times (a+(11-1)d)$

            $= \sf 3\times (a+10d)$

            $=\sf 3a+30d$

            $=\sf 3 \times -3d +30d$

            $=\sf -9d+30d$

            $=\bf 21d$

∴   L.H.S = R.H.S

Hence proved.