Question

In a certain A.P. the 24th term is twice the 10th term. Prove that the 72nd term is twice the 34th term.

Answer 1

Answer:

Given :  

24th term (a24) is twice the 10th term  

a24 = 2a10

a + (24 - 1) d = 2(a + (10 - 1)d

[an = a + (n – 1) d]

a + 23d = 2(a + 9d)

a + 23d = 2a + 18d

a - 2a = 18d - 23d

-a = - 5d

a = 5d …………..(1)

 

We have to prove that, 72th term is twice its 34 term

a72 = 2a34

a + (72 -1)d = 2(a +(34 - 1)d

a + 71d = 2(a + 33d)

5d + 71d = 2(5d + 33d)

76d = 2(38d)

76d = 76d

LHS = RHS

Hence,it is proved that 72th term is twice its 34 term.

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Answer 2

$\huge{\underline\pink{Question}}$

In a certain A.P. the 24th term is twice the 10th term. Prove that the 72nd term is twice the 34th term.

$\huge{\underline\pink{Solution}}$:-

Given,

24th term (a24) is twice the 10th term.

$\bold{a24 \:= \:2a10}$

2 + (24 - 1) d = 2 (a + (10 - 1) d

$\bold{[an \:= \:a \:+ \:(n \:- \:1) \:d]}$

a + 23d = 2 (a + 9d)

a + 23d = 2a + 18d

a - 2a = 18d - 23d

-a = -5d

$\bold{a \:= \:5d}$

$\bold{We \:have \:to \:prove \:that, \:73th \:term \:is \:twice \:its \:34 term.}$

a72 = 2a34

a + (72 - 1) d = 2 (a + (34 - 1) d

a + 71d = 2 (a + 33d)

5d + 71d = 2 (5d + 33d)

76d = 2 (38d)

$\bold{76d \:= \:76d}$

$\bold{L.H.S. \:= \:R.H.S}$

$\bold{Hence, \:it \:proved \:that \:72th \:term \:is \:twice \:its \:34 \:term.}$

$\huge{\mathfrak\pink{Hope \:it \:is \:helpful}}$