86. The sum of the first four terms of an A.P. is 56. The sum of the last four terms is 112. If its first term is 11, then find the number of terms.

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