Let 3, 6, 9, 12,…. upto 78 terms and 5, 9, 13, 17,…. upto 59 terms be two series. Then, the sum of the terms common to both the series is equal to…………

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, 7, 11, 15, 19, 23, 27, . 403 = AP₁

2, 5, 8, 11, 14, 17, 20, 23, . 401 = AP₂

so common terms A.P.

11, 23, 35, ., 395

->395 = 11 + (n – 1) 12

->395 – 11 = 12 (n – 1)

$\frac{384}{12}=n-1$    

32 = n – 1

n = 33

Sum =  $\frac{33}{2}$ [2×11+ (32)12]

=  $\frac{33}{2}$ [22 + 384]

= 6699

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

a – 3 = b – a