Question

first terms of two APs are 2 and 5 and their common differences is 3 and -2 respectively if the difference of the sums of n terms of these two AP is 195 find the number of terms solve it ​

Answer 1

Answer:

n = 10

Step-by-step explanation:

A = 2                                                        

D = 3                                                                          

No. of terms = n                                                        

∴ The A.P. is 2,5,8,11,...............

$S_n{}$ = $\frac{n}{2}[2A+(n-1)D]$

    = $\frac{n}{2}[2(2)+(n-1)3]$

    = $\frac{n}{2}[4+3n-3]$

    = $\frac{n}{2}[3n+1]$                           ------------------(i)

a = 5

d = -2

No. of terms = n

∴ The A.P. is 5,3,1,-1,-3,...............

$s_{n}$ = $\frac{n}{2}[2a+(n-1)d]$

    = $\frac{n}{2}[2(5)+ (n-1)(-2)]$

    = $\frac{n}{2}[10-2n+2]$

    = $\frac{n}{2}[12-2n]$                     -------------------------(ii)

   

ATQ(according to the question)

$S_{n}-s_{n}$ = 195

$[\frac{n}{2}(3n+1)] - [\frac{n}{2}(12-2n)] = 195$

$\frac{3n^{2} }{2}+\frac{n}{2}-\frac{12n}{2}+\frac{2n^{2} }{2} = 195\\\frac{5n^{2} }{2} -\frac{11n}{2} = 195$

$\frac{n}{2}(5n-11) = 195\\5n^{2} -11n = 390$

$5n^{2} -11n-390$ = 0

$5n^{2}-50n+39n-390 = 0\\ 5n(n-10)+39(n-10) = 0\\(5n+39)(n-10) = 0$

∴n-10 = 0                                &         5n+39 = 0

 n=10                                      &            n=$\frac{-39}{5}$

Because 'n' cannot in fraction form or with a negative sign.

Therefore n=10