Question

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

Answer 1

Ln = 7 -3n  ------(1)

l1 = 7- 3*1 = 7-3 =4

l2 = 7 - 3*2 = 7 - 6 =1

l3 = 7 - 3*3 = 7 - 9 = -2

first term = a = 4

common difference = d = l2 - l1 = 1-4 = -3

sum of 25 terms = sn = n/2[2a+(n-1)d]

s25 = 25/2 * [ 2*4 + (25-1)(-3)]

 = 25/2 * [8 + 24(-3)]

= 25/2 * [ 8- 72]

= 25/2 * [-64]

= - 25 *32

= -800

therefore

sum of 25 terms in an ap is -800

Answer 2

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

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

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

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

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

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

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

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

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