Question

For what value of n, are the nth terms of two APs:63, 65, 67,... and3, 10, 17,.... equal?

Answer 1

Hey there !!


Let the nth terms of the given progressions be $t \tiny{n}$ and $T \tiny{n}$ respectively.


▶ The first AP is 63, 65, 67...... .

Let its first term be a and common difference be d. Then,

=> a = 63 and D = ( 65 - 63 ) = 2.

So, its nth term is given by

$t \tiny{n}$ = a + ( n - 1 )d.

=> $t \tiny{n}$ = 63 + ( n - 1 ) × 2.

=> $t \tiny{n}$ = 63 + 2n - 2.

=> $t \tiny{n}$ = 2n + 61.........(1).


▶ The second AP is 3, 10, 17..... .

Let its first term be A and common difference be D. Then,

=> A = 3 and D = ( 10 - 3 ) = 7.

So , its nth term is given by

$T \tiny{n}$ = A + ( n - 1 )D.

=> $T \tiny{n}$ = 3 + ( n - 1 ) × 7.

=> $T \tiny{n}$ = 3 + 7n - 7.

=> $T \tiny{n}$ = 7n - 4..........(2).


▶ Now,

A/Q,

=> $t \tiny{n}$ = $T \tiny{n}$ .

=> 2n + 61 = 7n - 4.

=> 7n - 2n = 61 + 4.

=> 5n = 65.

=> n = $\frac{65}{5} .$

$\huge \boxed{ \boxed{ \bf => n = 13. }}$


✔✔ Hence, it is solved ✅✅.

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THANKS


#BeBrainly.

Answer 2

Answer:

hope it helps . please mark it as brainliest