Question

8 Firon what value of n are the nth
tesum of the followiny A139 APS
are some 13,19,25... and 69,689​

Answer 1

Answer:

it's the answer of Q11 from the attached image.

The AP is

17, 15, 13, 11, ...

So, first term, a = 17,

second term = 15.

Then, common difference,

d = 15 - 17 = - 2.

Let us consider the AP contains 'n' number of terms.

Sum of an AP series = 72

=> (n/2)[2a + (n-1)d] = 72

=> (n/2)[2×17 + (n-1)×(-2)] = 72

=> (n/2)[34 - 2n + 2] = 72

=> (n/2)(36 - 2n) = 72

=> n(18 - n) = 72

=> n² - 18n + 72 = 0

=> n² - 12n - 6n + 72 = 0

=> n(n - 12) - 6(n - 12) = 0

=> (n - 12)(n - 6) = 0

So, n = 12, 6.

Here, n = 6 is the required number of terms.

Since the AP is having first term 17 (positive) and common difference (-2) (negative), there are two values of n.

The sum of the last six terms (when n = 12) is 0 indeed.

So, only (6 - 4) = 2 or (12 - 4) = 8 terms need to be added.

Answer for the question you have written under the image..

Let the nth terms of the given progressions be tn and  Tn .Tn respectively.

• The first AP is 13, 19, 25...... .

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

=> a = 13 and D = ( 19 - 13 ) = 6.

So, its nth term is given by

tn  = a + ( n - 1 )d.

=> tn   = 13 + ( n - 1 ) × 6.

=>  tn = 13 + 6n - 6.

=>tn  = 6n + 7........(1).

• The second AP is 69, 68, 67..... .

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

=> A = 69 and D = ( 68 - 69 ) = -1.

So , its nth term is given by

Tn  = A + ( n - 1 )D.

=> Tn = 69 + ( n - 1 ) × (-1).

=> Tn  = 69 - n + 1.

=> Tn  = 70 - n..........(2).

• Now,

A/Q,

=> tn =Tn

=> 6n + 7 = 70 - n.

=> 6n + n = 70 - 7.

=> 7n = 63.

=> n = 63/7

=>n=9. 

these answers can have errors too...so please read these answers carefully...

hope it's gonna help..