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