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​The First and the Last Terms of an A.P. Are 7 and 49 Respectively. If Sum of All Its Terms is 420, Find Its Common Difference. - Mathematics

Sum

​The first and the last terms of an A.P. are 7 and 49 respectively. If sum of all its terms is 420, find its common difference. 

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Solution

Let a be the first term and d be the common difference.
We know that, sum of first n terms = Sn = \[\frac{n}{2}\][2a + (n − 1)d]
Also, nth term = an = a + (n − 1)d
According to the question,
a = 7, an = 49 and Sn = 420

Now,
an = a + (n − 1)d
 49 = 7 + (n − 1)d
​⇒ 42 = nd − d
​⇒
 nd − = 42                     ....(1)

Also,
Sn = \[\frac{n}{2}\][2 × 7 + (n − 1)d]
⇒ 420 = \[\frac{n}{2}\][14 + nd − d]
⇒ 840 = n[14 + 42]               [From (1)]
⇒ 56n = 840
⇒ n = 15                              ....(2) 
On substituting (2) in (1), we get
nd − = 42
⇒ (15 − 1)= 42
⇒ 14= 42
⇒ = 3

Thus, common difference of the given A.P. is 3.

 

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APPEARS IN

RD Sharma Class 10 Maths
Chapter 5 Arithmetic Progression
Exercise 5.6 | Q 26 | Page 52
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