Question

In an A.P. the first term is -5 and last term is 45. If sum of all numbers in the A.P. is 120, then how many terms are there? What is the common difference? Solve the word problem

Answer 1

$Solution$
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tn=a+(n-1)d=45

=>(n-1)d=50

=>d=(50)/(n-1).....1)

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now,

Sn=n/2{2a+(n-1)d}=120

=>n/2{2a+50}=120

=>n/2{40}=120

=>20n=120

=>n=6

now,

putting the value of n in eq1) we have

d=50/5

d=10

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$hope\\ it \\helps$

Answer 2

Your answer is --

Given, first term a = -5 and

last term l = 45

So,

Tn = 45

=> a + (n-1)d = 45

=> -5 + (n-1)d = 45

=> (n-1)d = 50 ......(1)

since, sum of all numbers in the A.P. is 120.

So,

Sn = n/2{ 2a + (n-1) d }

=> 120 = n/2 ( 2×-5+50)

=> 120 = n/2 × 40

=> n = 120/20

=> n = 6

therefore, there are 6 term in the AP

Now, put n = 6 in equation (1), we get

(6-1) d = 50

=> d = 50/5

=> d = 10

Hence, common difference is 10


【 Hope it helps you 】