The first term of an AP is -5 and the last term is 45. If the sum of the terms of the AP is 120, then find the number of terms and the common difference

Solution:

Given, the first term of an AP, a = -5

Last term of an AP, l = 45.

Sum of the term of the AP, S = 120

We have to find the number of terms and the common difference.

We know that, if l is the last term of an AP, then the sum of the terms is given by

S = n/2[a+l]

So, 120 = n/2[-5+45]

120 = n/2[40]

120 = 20n

n = 120/20

n = 6

The nth term of an AP is given by

aₙ = a + (n - 1)d

Given, a₆ = 45

45 = -5 + (6 - 1)d

45 + 5 = 5d

50 = 5d

d = 50/5

d = 10

Therefore, the number of terms is 6 and the common difference is 10.

✦ Try This: The first term of an AP is 5 and the last term is 45. If the sum of the terms of the AP is 400, then find the number of terms and the common difference

☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 5

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NCERT Exemplar Class 10 Maths Exercise 5.3 Problem 20

The first term of an AP is -5 and the last term is 45. If the sum of the terms of the AP is 120, then find the number of terms and the common difference

Summary:

The first term of an AP is -5 and the last term is 45. If the sum of the terms of the AP is 120, then the number of terms and the common difference are 6 and 10

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