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Step-by-Step Solution
Step 1: Express b and c in terms of a and r
Since a, b, and c are in Geometric Progression (G.P.) with first term a (nonzero) and common ratio r, we have:
b = ar
c = ar^2
Step 2: Write down the A.P. terms
We are given that 3a, 7b, and 15c are in Arithmetic Progression (A.P.). Thus:
3a, 7(ar), 15(ar^2)
Step 3: Use the A.P. property
For an A.P., the difference between consecutive terms is constant. Therefore,
(7b) – (3a) = (15c) – (7b)
Substitute b = ar, c = ar^2 to get:
7(ar) – 3a = 15(ar^2) – 7(ar)
It is often simpler to equate the middle term twice the middle term equals sum of first and third terms:
2 × (7ar) = 3a + 15(ar^2)
14ar = 3a + 15ar^2
Step 4: Solve for r
Rearrange the equation:
15r^2 – 14r + 3 = 0
We solve this quadratic equation for r. The solutions are:
r = 1/3, 3/5
But given
0 < r ≤ 1/2,
we must reject
r = 3/5 (since 3/5 = 0.6 > 1/2).
Therefore,
r = 1/3
is the acceptable common ratio.
Step 5: Determine the common difference
The common difference d of the A.P. is:
d = 7b – 3a = 7(ar) – 3a
Using
r = 1/3
gives
d = 7 × a × (1/3) – 3a = (7a)/3 – 3a = (7a – 9a)/3 = -(2a)/3
Step 6: Find the fourth term of the A.P.
The first three terms of the A.P. are 3a, 7b, 15c. The fourth term T4 is:
T4 = 15c + d
Since
c = a(1/3)^2 = a/9
and
d = -(2a)/3,
we get:
T4 = 15 (a/9) + (-(2a)/3)
= (15a)/9 – (2a)/3
= (5a)/3 – (2a)/3
= a
Final Answer
The fourth term of the A.P. is
a.
