If 3, 7, 11,...., 403 = AP₁

2, 5, 8, ...., 401 = AP₂

Find sum of common term of AP₁ and AP₂

First term = a

Common difference = d

Given: a + 5d = 2        . (1)

Product (P) = (a₁a₅a₄) = a (a + 4d) (a + 3d)

Using (1)

P = (2 – 5d) (2 – d) (2 – 2d)

-> = (2 – 5d) (2 –d) (– 2) + (2 – 5d) (2 – 2d) (– 1) + (– 5) (2 – d) (2 – 2d)

$\frac{dP}{dd}$ = –2 [ (d – 2) (5d – 2) + (d – 1) (5d – 2)

a, ar, ar², ….ar⁶³

a+ar+ar² +….+ar⁶³ = 7 [a + ar² + ar⁴ +.+ar⁶²]

$\frac{a\left(1-r^{64}\right)}{\left(1-r\right)}=7\frac{a\left(1-r^{64}\right)}{\left(1-r^2\right)}$            

1 + r = 7

r = 6

S₂₀ = $\frac{20}{2}$ [2a + 19d] = 790

2a + 19d = 79              . (1)

$S_{10}\frac{10}{2}\left[2a+9d\right]=145$

2a + 9d = 29                . (2)

from (1) and (2) a = –8, d = 5

$S_{15}-S_5=\frac{15}{2}\left[2a+14d\right]-\frac{5}{2}\left[2a+4d\right]$

$=\frac{15}{2}\left[-16+70\right]-\frac{5}{2}\left[-16+20\right]$  

= 405 – 10

= 395

3, a, b, c are in A.P.

a – 3 = b – a                                             

a + d, a + 7d and a + 43d are 1^st, 2^nd, 3^rd term of G.P.

$\frac{a+7d}{a+d}=\frac{a+43d}{a+7d}$              

⇒ (a + 7d)² = (a + d) (a + 43d)

⇒ a² + 49d² + 14d = a² + 44ad + 43d³

⇒ 6d² = 30ad

⇒ d² = 5d

⇒ d = 0, 5

a = 1, d = 5

  $S_{20}=\frac{20}{2}\left[2+\left(19\right)5\right]$          

= 10 [95 + 2]

= 970