Question

For what value of n, and nth terms of the arithmetic progressions 63, 65, 67, ...... and 3, 10, 17, ..... are 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 = $\frac{65}{5} .$

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

→ 13th term = a + 12d.

= 3 + 12 × 7.

= 3 + 84.

= 87.


✔✔ Hence, it is solved ✅✅.

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$\huge \boxed{ \mathbb{THANKS}}$


$\huge \bf{ \#BeBrainly. }$

Answer 2

Heya!!

For first AP: 63,65,67....

Here a = 63,
d = 65-63 = 2

For second AP: 3,10,17....

Here a' = 3
d' = 10-3 = 7

According to question ;

an = a'n

=) a + (n-1)d = a' + (n-1)d'

=) 63 + (n-1)2 = 3 + (n-1)7

=) 63 + 2n - 2 = 3 + 7n - 7

=) 61 +4 = 5n

=) 65 = 5n

=) 65/5 = n

=) n = 13

a13 = 3 + (13-1) 7

= 3+12(7)

= 3 + 84

= 87.

Hope it helps uh!