Question

how many terms of the A.p.24,21,18,... must be taken so that their sum is 78?​

Answer 1

Answer:

tn= a+(n-1)d

Explanation:

78=24+(n-1)-3

=-17

Answer 2

Answer:

13 terms

Explanation:

a (first term) = 24

d (common difference) = -3

Sum of 'n' number of terms = 78

Sum = n/2 (2a +(n-1)d)                              

78 = n/2 (2 * 24 + (n-1)*(-3))

=> 78 = n/2 (48 + (-3n +3))

=> 78 = n/2 (48-3n +3)

=> 78 = n/2 (51 - 3n)

=> 78*2 = n(51 - 3n)

=>$156 = 51n - 3n^{2}$

=> $3n^{2} -51n +156$ = 0

=> $n^{2} - 17n +52 = 0$

=> $n^{2} -13n - 4n +52 = 0$

=> n (n-13) - 4 (n-13) =0

=> (n-4)(n-13) =0

n-4=0                                     or                          n-13 =0

n=4                                        or                           n = 13

Sum = n/2 (a+l)  --------------------------------(i)

$t_{n} = a + (n-1)d$  --------------------------------(ii)

as according to verification by applying both the values of n in the equations (i) and (ii) ,it is found that n=13