Question

fing the sum of the first 25 terms of an AP whose nth term is given by an=7-3n

Answer 1

Answer:

-800

Step-by-step explanation:

Total no.of terms; n = 25

Given:

     $n^{th}$ term = 7-3n

For the first term, n=1

    1st term = 7-3*1 = 7-3

               a = 4

For the 25th term, n=25

       25th term = 7- 3*25 = 7-75

                  $a_{n}$ = -68

Sum of first 25 terms of the AP = n/2(a+$a_{n}$)

                                                       = 25/2(4+(-68)

                                                       = 25/2(4-68)

                                                       = 25/2 * (-64)

                                                       = 25*(-32)

                                                      = -800

∴ sum of first 25 terms of the given AP is -800.

HOPE THIS HELPZ :)

Answer 2

$\orange{ \rm \boxed{ \star \: given - }}$

$\boxed{ \rm \:a _{n} = (7n - 3n)}$

$\boxed{ \large \rm \to \: to \: find \: out \implies \: s_{25} = {?}}$

$\pink{ \boxed{ \rm \:solution:-}}$

$\rm \: t _{n} = (7n - 3n) \implies \: t _{1} = (7 - 3 \times 1) \: and \: t _{2} = (7 - 3 \times 2) = 1 \\ \rm \therefore \: a = 4 \: and \: d = (t _{2} - t _{1}) = (1 - 4) = - 3$

$\rm \: \therefore \: sum \: of \: 25 \: terms \: is \: given \: by \\ \rm s _{25} = \frac{n}{2} \times \large( \small \: 2a + (n - 1)d \large) \: \: where \: n = 25$

$= \large{ \frac{25}{2} } \times ( \small2 \times 4 + (25 - 1) \times ( - 3) \large) \: = \frac{25}{2 } \times ( - 64)$

$= 25 \times ( - 32) = - 800$

$\large \red{ \boxed{ \rm \: hence \: the \: sum \: of \: first \: 25 \: term \: is - 800}}$