Question

Question 2
If 8 times their 17th term of an A.P is equal to 6 times its 9th term, then the 41st term
of the A.P is?​

Answer 1

Answer:

T

n

=a+(n−1)d

Where a is the first term, d is the common difference and T

n

denotes the n-th term of an AP.

Given:

19=a+5d

41=a+16d

Solving the above equations gives: d=2,a=9

Therefore, T

40

=9+39⋅2=

87

Answer 2

Let  a be the first term be $T_n$  and $d$ be the common difference of the AP.

Then, nth term of the AP is given by

$T_n$ = a+(n−1)d

Given,

    8 × $T_1_7$ = 6 × $T_9$

​⇒ 8(a+(17−1)d)=6(a+(9−1)d)

⇒ 8(a+16d)=6(a+8d)

divide the equation by 2

⇒ 4(a+16d)=3(a+8d)

⇒ 4a+64d=3a+24d

⇒ 4a-3a=24d-64d

⇒ a = −40d

So,  

$T_4_1$ = a + (41 − 1) d

     = −40d+40d

     = 0

Therefore, the 41st term of the A.P is 0

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