Question

Find the sum of the first 25 tearms of an AP whose nth term is given by an = 7-3n.

Answer 1

Step-by-step explanation:

an=7-3n

a1=7-3=4

a2=7-6=1

d=a2-a1=1-4=-3

Sn=n/2[2a+(n-1)d]

S25=25/2[2(4)+(25-1)(-3)]

S25=25/2[8+24(-3)]

S25=25/2(64)

S25=25*32=800

hope this will help you....

Answer 2

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

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

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

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

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

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

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

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

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