Question

if the ratio of 19th term to 22nd term of an AP is 22:29,then ratio of sum it's 7 terms to the sum of it's 10 terms will be

Answer 1

In AP, we have

a(19) = a + (19-1)d = a + 18d

a(22) = a + (22-1)d = a + 21d
So, we have the ratio as
(a + 18d)/(a + 21d) = 22/29
29(a + 18d) = 22(a + 21d)
29a + 522d = 22a + 462d
7a = -60d
a = -60d/7
Sum of n terms is given by
S(n) = n/2 [ 2a + (n-1)d ]
S(7) = 7/2 [ 2a + 6d ]
S(10) = 10/2 [ 2a + 9d ]
S(7) / S(10) is given by
7 [2a + 6d ] / 10 [ 2a + 9d ]
(14a + 42d) / (20a + 90d)
Putting value of a, we get
[ 14(-60d/7) + 42d ] / [ 20(-60d/7) + 90d ]
[ -120d + 42d ] / [ -1200d/7 + 90d ]
[ -78d ] / [ -1200d/7 + 630d/7 ]
[ -78d ] / [ -570d/7 ]
= [ 78*7 ] / 570
= 13*7 / 95 = 91 / 95
Therefore, the required ratio is 91 : 95

Answer 2

Given conditions ⇒

t₁₉ : t₂₂ = 22 : 29

Thus, 
[a + (19 - 1)d] × 29 = [a + (22 - 1)d] × 22
⇒ 29a + 522d = 22a + 462d
⇒ 29a - 22a = 462d - 522d
∴ 7a = - 60d
∴ a = -60d/7

Therefore, the value of a = -60d/7.

Sn = n/2[2a + (n - 1)d]
∴ S₇ = 7/2 [ 2(-60d/7) + 6d]
 ∴ S₇ = -60d + 21d= -39d

Now, Let us find the Sum of the First 10 term of the Sequence

 S₁₀ = 10/2 [2(-60d/7) + (10 - 1)d]
 = 5[-120d/7 + 9d]
 = 5[(-120d + 63d)/7]
 = 5[-57d/7]= -285d/7

Thus,
 S₇/S₁₀ = -39d/(-285d/7) 
 = (39 × 7)/285
 = 273/285
 = 91/95

Hence, the Ratio of the sum of 7 term to the sum of the 10th term is 91 : 95.


Hope it helps.