Question

Go In an AP. the first term is -5 and the last
form is 45. If the sum of all numbers in the
A-
Pjs 120, then bow many terms are there?
What is the common difference?​

Answer 1

Answer:

Given:

1. First term, a= -5
2. Last term, $a_{n}$ = 45
3. Sum of all the terms, $S_{n}$= 120

To find : The number of terms

Proof:

Let the number of terms= n

According to the question,

$S_{n}$ = $\frac{n}{2}$ ( a+ $a_{n}$ )

Substituting the values, we get:

= 120 =  $\frac{n}{2}$ ( -5 +45 )

⇒ 120 =  $\frac{n}{2}$ (40)

⇒ 120 = n (20)

⇒ n = $\frac{120}{20}$

⇒ n = 6

∴ Number of terms in the given A.P. = n = 6

Now,

$a_{n}$ = 45                                         (Given)

But,

$a_{n}$ = a +(n-1) d

Where

1. a= first term
2. $a_{n}$ = last term
3. n= total number of terms
4. d = common difference.

Substituting the values in the equation, we get:

= 45 = -5 + (6-1) d

⇒ 45 +5 = 5d

⇒ 50 = 5d

⇒ d= $\frac{50}{5}$

⇒ d = 10

∴ The common difference = d= 10

Hence, number of terms = n= 6

and the common difference = d = 10

Proved.

Hope you got that.

Thank You.