AP 7th term : a + 6d
AP 3rd term : a + 2d
AP 1st term : a

GP 1st term : b
GP 2nd term : br
GP 3rd term : br²

so a = br² ①
a + 2d = br ②
a + 6d = b ③

Sub ① into ②,

br² + 2d = br
2d = br - br² ④

Sub ④ and ① into ③,

br² + 3(br - br²) = b
3br - 2br² = b

2br² - 3br + b= 0
b(2r² - 3r + 1) = 0
b(2r - 1)(r - 1) = 0

b = 0 or r = ½ or r = 1

If r = 1, it would imply b = br = br² which means a + 6d = a + 2d = a, which implies d = 0. But question says common difference is non-zero. So r ≠ 1.

Likewise, if b = 0, then b = br = br² = 0, which implies that a + 6d = a + 2d = a
= 0. Then 6d = 2d, which implies d = 0.

But d ≠ 0. So b ≠ 0. (And if you had b = 0, you won't be having a G.P at all)


So r = ½. Since |r| < 1, series is convergent




So from ①, a = b(½)² = ¼b
From ④, 2d = b(½) - b(½)² = ¼b

So 2d = a , which means d = ½a

So 7th term = a + 6d = a + 3(a) = 4a
3rd term = a + 2d = a + a = 2a

Now, question says 7th term is 3.
So 4a = 3
a = ¾

Therefore ¼b = ¾
b = 3
d = ½(¾) = ⅜




Sum of first n terms of AP - sum of first n terms of GP ≥ 100


n/2 (2a + (n-1)d) - b(1 - rⁿ)/(1 - r) ≥ 100

sub all values in,

n/2 ( 2(¾) + (n-1)(⅜) ) - 3(1 - ½ⁿ)/(1 - ½) ≥ 100


¾ n + 3/16 n² - 3/16 n - 6 + 6(½ⁿ) ≥ 100

3/16 n² + 9/16 n - 6 + 6/2ⁿ ≥ 100

Multiply both sides by 16,

3n² + 9n - 96 + 96/2ⁿ ≥ 1600
3n² + 9n + 96/2ⁿ ≥ 1696


Solving for n using GC,

n ≥ 22.324 or n ≤ -4.13099 (N.A, number of terms cannot be negative)

Since n is an integer,
smallest value of n = 23