RD Chapter 19 Arithmetic Progressions Ex 19.2 Solutions
Question - 11 : - The 10th and 18th term of an A.P. are 41and 73 respectively, find 26th term.
Answer - 11 : -
Given:
10th termof an A.P is 41, and 18th terms of an A.P. is 73
So, a10 =41 and a18 = 73
We know, an =a + (n – 1) d [where a is first term or a1 and d is the commondifference and n is any natural number]
When n = 10:
a10 =a + (10 – 1)d
= a + 9d
When n = 18:
a18 =a + (18 – 1)d
= a + 17d
According to question:
a10 =41 and a18 = 73
a + 9d = 41 ………………(i)
And a + 17d =73…………..(ii)
Let us subtractequation (i) from (ii) we get,
a + 17d – (a + 9d) =73 – 41
a + 17d – a – 9d = 32
8d = 32
d = 32/8
d = 4
Put the value of d inequation (i) we get,
a + 9(4) = 41
a + 36 = 41
a = 41 – 36
a = 5
we know, an =a + (n – 1)d
a26 =a + (26 – 1)d
= a + 25d
Now put the value of a= 5 and d = 4 in a26
a26 =5 + 25(4)
= 5 + 100
= 105
Hence, 26th termof the given A.P. is 105.
Question - 12 : - In a certain A.P. the 24th term is twice the 10th term.Prove that the 72nd term is twice the 34th term.
Answer - 12 : -
Given:
24th termis twice the 10th term
So, a24 =2a10
We need to prove: a72 =2a34
We know, an =a + (n – 1) d [where a is first term or a1 and d is common differenceand n is any natural number]
When n = 10:
a10 =a + (10 – 1)d
= a + 9d
When n = 24:
a24 =a + (24 – 1)d
= a + 23d
When n = 34:
a34 =a + (34 – 1)d
= a + 33d ………(i)
When n = 72:
a72 =a + (72 – 1)d
= a + 71d
According to question:
a24 =2a10
a + 23d = 2(a + 9d)
a + 23d = 2a + 18d
a – 2a + 23d – 18d = 0
-a + 5d = 0
a = 5d
Now, a72 =a + 71d
a72 =5d + 71d
= 76d
= 10d + 66d
= 2(5d + 33d)
= 2(a + 33d)[since, a = 5d]
a72 =2a34 (From (i))
Hence Proved.