║⊕QUESTION⊕║
Mathematics is a place where you can do things that you can't do in the real world.
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CLASS 11
SEQUENCES AND SERIES
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The sum of n terms of two A.P's are in the ratio (2n+4):(3n+5). Find the ratio of their 15th terms
Answers
Step-by-step explanation:
the ratio of 15 th terms of the AP are 66:67
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The solution given above is right but have one problem only I.e 15 term of an AP written as a + 14d not as a+15d
◆【Solution】◆
I want to solve this question in two parts means with two different APs.
Sum of n digits of an A.P with first term a
common difference d can be written as...
●【S = (n/2)(2a + {n-1}d)】●
Acc to question
(n/2)(2a + {n-1}d) = (2n + 4)k
where k is an arbitrary constant.
(n/4)(a + {n-1}d/2) = (2n + 4)k...(1)
Let the first term of second AP be A
Common difference be D
Acc to question....Sum of this AP is ....
(n/2)(2A + {n-1}D) = (3n+5)k
(n/4)(A + {n-1}D/2) = (3n+5)k....(2)
For first and second AP nth term is
a + {n-1}d
15th term can be written as....
a + 14d.....
but if we replace d by d/2 than 15th term can be written as....
a + {n-1}d/2
where n is 29....
So we need to replace n by 29 and divide both equation to eliminate k.
In first equation.....
(29/4)(a + 14d) = 62K
In second equation
(29/4)(A + 14D) = 92K
we know that a + 14d is the 15th term of first AP
And A + 14D is 15th term of second AP
Divide both...
◆【t(15) : T(15) = 31 : 46】◆
Sorry for any Mathematical mistake but this is the way you solve this type of questions.
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