Question

The sums of n terms of two Arthematic progression are in the ratio (7n-5):(5n+17). Show that their 6th term are equal?

Answer 1

Answer:

The sixth term of both APs is 36

Step-by-step explanation:

Let a, d and b, e be the first term and the common difference of the first and the second AP respectively.

∴ [(n/2)[2a+(n-1)d]/[(n/2)[2b+(n-1)e] is their ratio of n terms which is equal to (7n-5)/(5n+17) ........ (given)

Now (7n-5)/(5n+17) = (7n-7+7-5)/(5n-5+5+17)

(adding 0 to both numerator & denominator)

= [7(n-1)+2]/[5(n-1)+22]

= [2×1+7(n-1)]/[2×11+5(n-1)] =  [(n/2)[2a+(n-1)d]/[(n/2)[2b+(n-1)e] ....(see above)

∴ [2a+(n-1)d]/[(n/2)[2b+(n-1)e] =  [2×1+7(n-1)]/[2×11+5(n-1)]

Hence, a=1, d=7 and b=11, e=5

For AP1, a6=1+(6-1)7 = 1+35 = 36

For AP2, b6=11+(6-1)5 = 11+25 = 36.

Therefore, the sixth term of both the APs are equal.

See attachment for understanding easily