# Important Questions for Chapter 5 - Arithmetic Progressions

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**Important questions based on NCERT syllabus for Chapter 5 - Arithmetic Progressions:**

*Question 1*: If the ratio of sums to n terms of two AP's is (5n+3) : (3n+4), then what is the ratio of their 17th terms?

*Solution*:

According to the given condition.

(n/2[2a+(n−1)d]) / (n/2[2A+(n−1)D]) = 5n+3/3n+4

⇒ (a+(n−1/2)d) / (A+(n−1/2)D) = 5n+3/3n+4

For the ratio of the 17th terms,

(n−1)/2 = 16

n = 33

⇒ (a+16d) / (A+16D) = (5(33)+3) / (3(33)+4)

= 168/103

= 168:103

*Question 2*: In an A.P. the sum of first n terms is 3n^2/2 + 13n/2. Find the 25th term.

*Solution*:

Sn = 3n^2 + 13n^2

a_{n} = Sn−Sn−1

a_{25}=S_{25}−S_{24}

a_{25} = 3(25)^2 + 13(25)/2 − 3(24)^2 + 13(24)/2

a_{25} = 1/2[3(25^2−24^2) + 13(25−24)]

a_{25} = 1/2[3(49)+13]

a_{25} = 80

*Question 3*: The sum of n terms of an AP is 3n2+5n. The mth term of the AP is 164. Find the value of m.

*Solution*:

We will find an expression for the nth term. To do that, we will find the difference between the sum of n terms and the sum of (n-1) terms:

Tn = Sn−Sn−1

=(3n^2+5n) − {3(n−1)^2 + 5(n−1)}

=3{n^2 −(n−1)^2} + 5{n−(n−1)}

=3(2n−1) + 5 = 6n + 2

Since the mth term is 164, we have:

Tm = 6m + 2 = 164

⇒ m = 27