Question

.1 Find the mth term of an Arithmetic progression whose 12th

term exceeds the 5th term by 14 and the sum of both terms

is 36​

Answer 1

✬ mth Term = 1 + 2m ✬

Step-by-step explanation:

Given:

• 12th term of AP exceeds 5th term by 14.
• Sum of both terms is 36.

To Find:

• What is the mth term of AP ?

Solution: As we know that , nth term of AP series is given by

★ aⁿ = a + (n – 1)d ★

[ 12th term of AP will be ]

➟ a¹² = a + (12 – 1)d

➟ a + 11d

[ 5th term of AP will be ]

➟ a⁵ = a + (5 – 1)d

➟ a + 4d

• Sum = a + 11d + a + 4d = 36......i

A/q

• 12th term of AP exceeds 5th term by 14.

➮ a + 11d = a + 4d + 14

➮ a + 11d – (a + 4d) = 14

➮ a + 11d – a – 4d = 14

➮ 7d = 14

➮ d = 14/7

➮ d = 2

Now put the value of d in equation i .

➯ a + 11(2) + a + 4(2) = 36

➯ a + 22 + a + 8 = 36

➯ 2a = 36 – 30

➯ a = 6/2

➯ a = 3

So we have

• First term = a = 3
• Common difference = d = 2

Therefore, mth term will be

➯ am = a + (m – 1)d

➯ 3 + (m – 1)2

➯ 3 + (2m – 2)

➯ 1 + 2m

Hence, mth term of AP will be 2m + 1 or 1 + 2m.

Answer 2

♧Answer♧

• 12th term=a+11d

• 5th term=a+4d

acc to ques.

(a+11d)-(a+4d) = 14

7d = 14

d = 2

also (a+11d)+(a+4d) = 36

Putting the value of d and solving for a we get:

a = 3

• mth term

= a+(m-1)d

= 3+(m-1)2

mth term = 1+2m

✌