Question

find the sum of first 30 terms of AP whose nth term 2-3n​

Answer 1

Answer:

ANSWER= -1335

Step-by-step explanation:

$nth \: term = 2 - 3n \\ put \: the \: value \: of \: n = 1 \\ 2 - 3(1) = - 1 \\ put \: the \: value \: of \: n = 2 \\ 2 - 3(2) = - 4 \\ put \: the \: value \: of \: n =3 \\ 2 - 3(3) = - 7 \\ put \: the \: value \: of \: n =4 \\ 2 - 3(4) = - 10 \\ it \: is \: written \: as \: = \: ( - 1)( - 4)( - 7)( - 10)........n \: terms \\ a = - 1 \\ d = b - a = - 4 - ( - 1) \\ - 4 + 1 = - 3 = d \\ \\ to \: find \: sum \: of \: 30 \: terms \\ \\ formula \: of \: sum \: of \: n \: terms \\ sn = \frac{n}{2}(2a + (n - 1)d \\ \\ s30 = \frac{30}{2}(2( - 1) + (30 - 1) - 3 \\ 15( - 2 + 29( - 3)) \\ 15( - 2 + ( - 87)) \\ 15( - 89) \\ = - 1335 = answer$

Answer 2

QUESTION:

find the sum of first 30 terms of AP whose nth term 2-3n

FORMULA USED :

There are two formula to find the sum.

1.

$\huge\blue {s(n) = \frac{n}{2} [2a + (n - 1)d]}$

where;

a = first term

d =common difference

2.

$\huge\pink {s(n) = \frac{n}{2} (a + l)}$

where;

a = first term

l = last term

now come to main question;

nth term = 2 -3n

If we put the n =1 then we get the first term;

$n(1) = 2 - 3 \times 1$

$n(1) = 2 - 3 \\ n(1) = - 1$

$first \: term = - 1$

if we put the n = 2 then we get the second term;

$n(2) = 2 - 3 \times 2 \\ n(2) = 2 - 6 \\ n(2) = - 4$

$second \: term = - 4$

now;

we have to find common difference(d)

$\huge\pink {d = second \: term - first \: term}$

so,

$d = - 4 - ( - 1) \\ d = - 4 + 1$

$\huge\blue {d = - 3}$

now,

$\purple {s(30) = \frac{30}{2} [2 \times ( - 1) + (30 - 1) - 3]}$

$s(30) = 15 \times [ - 2 + 29 \times ( - 3)]$

$s(30) = 15[ - 2 + (-87)]$

$s(30) = 15 \times×(-89)$

$s(30) = - 1335$

FINAL ANSWER :

$\huge\pink {s(30) = - 1335}$