Question

A)the sixth term of an arithmetic progression is 23 and the sum of first ten terms is 200.Find the seventh term.

b.A geometric progression has first term 1 and common ratio r. A second geometric progression has first term 4 and common ratio r/4 . The two progressions have the same sum to infinity, s. Find the value of r and s.

Answer 1

Answer:

T₇ = 29

r = 4/5

s = 5

Step-by-step explanation:

Hi,

In an Arithmetic Progression,

Sum of r th term from the start and the end is constant.

In the first 10 terms of A.P, T₁+ T₁₀  = T₂ + T₉ = T₃ + T₈= ....

Sum of first ten terms = 5 * Sum of 5 th term and 6 th term

200 = 5( T₅ + T₆)

T₅ + T₆ = 40

T₅ = 40 - 23 = 17

T₆ = 23

Common difference = d = T₆ - T₅

d = 6

Seventh Term , T₇ = T₆ + d

T₇ = 23 + 6 = 29.

b) Geometric Progression 1

First Term = 1

Common Ratio = r

Sum of infinte terms, S = 1/1 - r

Geometric Progression 1

First Term = 1

Common Ratio = r

Sum of infinte terms, S₁ = 1/1 - r

Geometric Progression 2

First Term = 4

Common Ratio = r/4

Sum of infinte terms, S = 4/1 - r/4

Since Sum to infinite terms are equal, 1/ 1 - r = 4/ 1 - r/4

1 - r/4 = 4 - 4r

15r/4 = 3

r = 4/5.

Sum, S = 1/1 - 4/5 = 1/ 1/5  = 5

Hence, S = 5.

Hope, it helps !