Question

Please answer my question quickly I need it...
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Answer 1

hope this will help you :)

Answer 2

here,

$sn = 4n - {n}^{2}$
$sn - 1 = 4(n - 1) - ({n - 1})^{2}$

Sn-1=(4n-4) - (n^2 - 2n + 1)

Sn-1'= 4n-4-n^2+2n-1

Sn-1'= -n^2+6n-5


we know that......


$tn = sn - sn - 1$
tn = (4n -n^2) - (- n^2+6n-5)

tn = 4n - n^2 +n^2 - 6n + 5

tn = 4n -6n+5

tn = -2n+5

therefore,
the nth term is -2n +5

for 1st term put n=1

t1 = -2(1)+5
t1 = -2+5
t1 = 3

now, put n=2

t2 = -2(2)+5
t2=-4+5
t2 = 1

we have to find sum of first two terms

$sn = 4n - {n}^{2}$
$s2 = 4(2) - \: {2}^{2}$

$s2 \: = 8 - 4$

$s2 = 4$
for 3 rd term put n=3
in the nth term.

tn = -2n+5

t3 = -2(3)+5
t3 = -6+5
t3 = -1

similarly
for 10th term
put n=10
t10 = -2n+5
t10=-2(10)+5
t10=-20+5
t10=-15

THEREFORE,
THE FIRST TERM IS 3
second term is 1
sum of two terms is 4
3rd term is -1
tenth term is -15
nth term is -2n+5

$hope \: this \: helps$