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**Short Answer Type Questions**

**Q.1. If S _{n}, the sum of the first n terms of an AP is given by S_{n} = 2n^{2} 2+n, then find its n^{th} term. [CBSE 2019]**

Ans:

∴ a

Let nth term of the given AP be 120 more than its 21st term. Then,

a

Hence, the 31st term of the given AP is 120 more than its 21st term.

∴ t

Hence, t

Ans: Let a = first term,

Given, d=- 4,a

Ans:

Here, 2nd term - 1st term = 3rd term - 2nd term

⇒ 6 - 3 = 9 - 6 = 3

Since, common difference is same. Therefore, the above terms are in AP.

Here, a = 3, d = 6 - 3 = 3, n = 8

We know, the sum of n terms of an AP

where a = first term, d = common difference

Ans:

nth term of the AP is = a

(5 + 9 + 13 + ... 4- 81) 4- (-41 + (-39) + (-37) + ... + (-5) + (-3))

For the series 5 + 9 + 1 3 + ... 81

a = 5, d = 4 and a

nth term = a + (n - l)d = a

⇒ 5 ( n - 1)4 = 81

⇒ 4n = 80

⇒ n = 20

Sum of 20 terms for this series.

...(i)

Sum of 20 terms for this series

...(ii)

By adding (i) and (ii), we get

Sum of series = 860 - 440

= 420

According to question,

Thus, the sum of 12 terms of the given AP is 636.

...(i)

Now, putting the value of n in equation (i), we have

Hence, the number of terms is 16 and the common difference is 8/3.

Since, 216-208 = 224-216

Therefore, the given integers are in A.P.

Let first term (a) = 208,

common difference (d) = 224 -216 = 8,

Number of terms(n) = ?

Last term (a

We know that, a

⇒ 496 = 208 + (n - 1)(8)

⇒ 496 - 208 = (n - 1)(8)

⇒ 288 = 8n - 8

⇒ 8n = 288 + 8 = 296

⇒

Therefore, 37 integers are there between 200 and 500 divisible by 8.

In this progression, first term is 20

Also, second term - first term

third term - second term =

fourth term - third term =

∵ Difference between constructive term is constant,

therefore this progression is an AP.

Let nth term be the first negative term

∴

⇒

Now A.T.Q a

⇒

⇒

⇒

⇒ 83 - 3n < 0

⇒ 3n > 83

⇒

⇒

⇒ n

[ ∵ n is a natural number]

Thus, 28th term of the given progression is the first negative term.

**[CBSE Delhi 2017 (C)]****Ans:** Here a = 12, d = 4, a_{n} = 96

The formula is a_{n} = a + (n - 1 )d

Therefore 96 = 12 + (n - 1) x 4

Apply the formula for sum,

Hence, S_{22} = 11[24 + 21 x 4] = 11 [24 + 84]

= 11 x 108 = 1188.**Q.21. The ratio of the sum of first n terms of two different AP’s is (7n + 1) : (4n + 27), find the ratio of their mth terms. [CBSE (AI) 2016]****Ans:****Q.22. Find the 9th term from the end (towards the first term) of the AP 5, 9, 13,..., 185. [CBSE (Delhi) 2016]****Ans:** l = 185, d = 4

l_{9} = l - (n - 1) d = 185 - 8 x 4 = 153**Q.23. How many terms of the AP 18, 16, 14,.... be taken so that their sum is zero? [CBSE Delhi 2016]****Ans:****Q.24. The 4th term of an AP is zero. Prove that the 25th term of the AP is three times its 11th term. [CBSE (Al) 2016]****Ans:****Q.25. If the ratio of the sum of the first m and n terms of an AP is m ^{2}: n^{2}, show that the ratio of its m^{th} and n^{th} terms is (2m - 1) : (2n - 1). [CBSE (F) 2016]**

Let a

so

We need to find the values of 'n' and 'd'

We know that ...(i)

Substituting the values in (i)

⇒

⇒ n = 16

Also, a

⇒ 45 = 5 + (16 - 1 )d

⇒ 45 - 5 - 15d

⇒

⇒

Hence, there are 16 terms and the common difference is 8/3.

**Long Answer Type Questions**

**Q.1. The first term of an AP is 3, the last term is 83 and the sum of all its terms is 903. Find the number of terms and the common difference of the AP. **

**[CBSE 2019 ]****Ans:** We have an AP, in which

a = 3, last term, a_{n} = l = 83

...(i)

No. of terms =21 and common deference = 4**Q.2. Which term of the Arithmetic Progression - 7, - 12 , - 17 , - 22 , ... will be - 82. Is - 100 any term of the A.P.? Give reason for your answer. [CBSE, Allahabad 2019]****Ans:****Q.3. The sum of four consecutive numbers in an AP is 32 and the ratio of the product of the first and the last term to the product of two middle terms is 7: 15. Find the numbers. [CBSE 2018]Ans: **Let the four consecutive numbers of the AP are

Therefore, four consecutive terms of the AP’s are 2, 6, 10, 14 or 14, 10, 6, 2 respectively.

...(i)

Hence S

Ans:

Then the sums of their n terms is given by

and

A.T.Q.

Ans:

a

Ans:

...(i)

15th term from the last = (50-15 + l)

Putting the value of d in (ii)

So, the AP formed is 3, 7, 11, 1 5 , ..... and 199.

Total distance covered by thief = 100 n metres

Total distance covered by policeman = 100 + 110 + 120 + ... + (n - 1) terms

Policeman took (n - 1) = ( 6 - 1 ) = 5 minutes to catch the thief.

The numbers of the houses are in AP, where a = 1 and d = 1

Sum of n terms of an AP =

Let X

Sum of number of houses preceding X

The sum of numbers of houses following Xth house is equal to S

Now, we are given that sum of number of houses before X is equal to the sum of the number of houses after X.

i.e., S

Since number of houses is positive integer, ∴ X = 35

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