In a certain $A . P .$ the $24^{t h}$ term is twice the $10^{t h}$ term. Prove that the $72^{n d}$ term is twice the $34^{t h}$ term.

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Solution

The $n$ th term of an A.P is given by
$a_{n} = a_{1} + (n - 1) d \dots . . e q (1)$
where $a_{1}$ is the first term of A.P
and d is the common difference
hence 10th term of an A.P is
$a_{10} = a_{1} + (10 - 1) d$
$a_{10} = a_{1} + 9 d \dots . . . e q (2)$
and 24th term is $a_{24}$
$a_{24} = a_{1} + (24 - 1) d$
$a_{24} = a_{1} + 23 d$
given that the 24th term is twice the 10th term
$⟹ a_{24} = 2 \times a_{10}$
$⟹ 2 \times a_{10} = a_{1} + 23 d \dots . . e q (3)$
put value of $a_{10}$ from eq(2)
$⟹ 2 \times (a_{1} + 9 d) = a_{1} + 23 d$
$⟹ 2 a_{1} + 18 d = a_{1} + 23 d$
$⟹ 2 a_{1} - a_{1} = 23 d - 18 d$
$⟹ a_{1} = 5 d . . . . . e q (4)$

now 34th term of A.P is given by
$a_{34} = a_{1} + (34 - 1) d$
put value of $a_{1}$ from eq(4) in above equation.
$a_{34} = 5 d + (33) d$
$a_{34} = 38 d \dots . . . e q (5)$

now 72nd term of A.P is given by
$a_{72} = a_{1} + (72 - 1) d$
put value of $a_{1}$ from eq(4) in above equation.
$a_{72} = 5 d + (71) d$
$a_{72} = 76 d$
$a_{72} = 2 \times 38 d$
we know that from eq(5)
$a_{34} = 38 d$
hence, $a_{72} = 2 \times a_{34}$

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