Math, asked by Vamprixussa, 10 months ago

║⊕QUESTION⊕║
Mathematics is a place where you can do things that you can't do in the real world.

CLASS 11
SEQUENCES AND SERIES

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

Answered by vermarishita
2

Step-by-step explanation:

the ratio of 15 th terms of the AP are 66:67

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Attachments:
Answered by anu24239
1

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.

#answerwithquality

#BAL

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