103. A person writes a letter to four of his friends. He asks each one of them to copy the letter and mail to four different persons with instruction that they move the chain similarly. Assuming that the chain is not broken and that it costs 50 paise to mail one letter. Find the amount spent on the postage when 8th set of letter is mailed.

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