CBSE NCERT Solutions Chapter 5 Arithmetic Progressions Exercise 5.3 Question 10
10. Show that
form an AP where
is defined as below:
(i)
(ii)
Also. find the sum of the first 15 terms in each case.
Solution (i)
We need to show that
form an AP where
Lets calculate values of
using
.
So, the sequence is of the form
Lets check difference between consecutive terms of this sequence.
11- 7 = 4
15- 11 = 4
19 – 15 = 4
Therefore, the difference between consecutive terms is constant which means terms
form an AP.
We have sequence
First term = a =7
Commom difference = d = 4
Applying formula,
to find sum of n terms of AP , we get
Therefore, sum of first 15 terms of AP is equal to 525.
Solution (ii)
We need to show that
form an AP where
Lets calculate values of
using
.
So, the sequence is of the form
Lets check difference between consecutive terms of this sequence.
-1-(4) = -5
-6-(-1) = -6 + 1 = -5
-11-(-6) = -11 + 6 = -5
Therefore, the difference between consecutive terms is constant which means terms
form an AP.
We have sequence
First term = a =4
Commom difference = d = -5
Applying formula,
to find sum of n terms of AP , we get
Therefore, sum of first 15 terms of AP is equal to -465.
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