Question

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

Answer 1

Hii mate..!!

Here is ur answer..!!

Answer: tn = 7-3n

if n= 1

t1 = 7-3(1)

t1 = 4

if n= 2

t2 = 7-3(2)

t2 = 1

∴ sequence = 4 , 1.......

a = 4

d = 1 - 4 => -3

an = a + (n-1) d

a25 = 4 + ( 25 - 1 ) -3

a25 = 4 + ( 24 ) -3

a25 = 4 - 72

a25 = 68

∴ 25th term of an ap = 68

I HOPE THIS MAY HELP U :)

Answer 2

Required answer:--

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

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

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

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

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

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

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

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

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