Question

The last term of an A.P. is 120. It's first term and common difference are 20 and 5

respectively. Find the sum of the A.P​

Answer 1

Answer:

first term = a = -5

last term = an = 45

Let the number of terms = n

Sum of n terms = 120

Sn = 120

n/2 [ a + an ] = 120

n/2 [ -5 + 45 ] = 120

n/2 ×( 40 ) = 120

n × 20 = 120

n = 120/20

n = 6

Let the common difference = d

a + ( n - 1 ) d = 45

-5 + ( 6 - 1 ) d = 45

-5 + 5d = 45

5d = 45 + 5

5d = 50

d = 10

Therefore ,

common difference= d = 10 ,

number of terms = n = 6

I hope this helps you.

Answer 2

Step-by-step explanation:

The nth––– term is

an=a1+(n−1)d.

The first term is a1=−5 and the last term is an=45.

The sum of the A. P. is Sn=120.

There are two formulas for the sum of an A. P. and they are:

Sn=n(a1+an)2 and

Sn=n(2a1+(n−1)d)2

where d is the common difference.

The first formula listed yields the value of n:

120=n(−5+45)2

⟹2(120)=n(40)

⟹240=40n

⟹24040=40n40

⟹6=n.

∴ there are 6 terms.

Next find d with the second formula for Sn , now known to be S6:

120=6(2(−5)+(6−1)d)2

⟹2(120)6=−5+5d

⟹40=−10+5d

⟹50=5d

⟹505=5d5

⟹10=d.

Check these answers:

a1=−5,

a2=−5+(2−1)(10)

=−5+10=5 and

a6=−5+(6−1)(10)

=−5+50=45✓