Q. 34.0( 282 Votes )

# In an AP:

(i) givenfind and

(ii) givenfind and

(iii) given find and

(iv) givenfind and

(v) givenfind a and

(vi) givenfind and

(vii) given find and

(viii) given find and

(ix) given find

(x) givenand there are total 9 terms. Find

Answer :

(i) To find: n, S_{n}Given:

a_{n} = 50

a = 5

d = 3

We know, nth term of an AP is given by

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

where a and d are first term and common difference respectively and n are the number of terms of the A.P

Number of terms can be calculated as follows:

a_{n} = a + (n – 1)d

Or,

50 = 5 + (n – 1)3

Or, (n – 1)3 = 50 – 5 = 45

Or, n – 1 = 15

Or, **n = 16**

Sum of N terms can be given as follows:

S_{16}

= 8(10 + 45)

= 8 x 55

= 440

**S**

_{n}= 440(ii) **To find: d and S _{13}**

**Given:**

**a = 7**

**a**

_{13}= 35

We know, nth term of an AP is

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

where a and d are first term and common difference respectively.

So, Common difference can be calculated as follows:

a_{n} = a + (n – 1)d

Or, 35 = 7 + 12d

Or, 12d = 35 – 7 = 28

Or, **d = 7/3**

Sum of n terms can be given as follows:

S_{13}

= (14 + 28)

=

=273**S _{13} = 273**

(iii) **To Find: a, S _{12}**

**Given:**

**d = 3**

**a**

_{12}= 37We know, nth term of an AP is

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

where a and d are first term and common difference respectively.

Therefore, First term can be calculated as follows:

a_{n} = a + (n – 1)d

Or, 37 = a + 11 x 3

Or, a = 37 – 33 = 4

**a = 4**

Sum of n terms can be given as follows:

S_{12}

= 6(8 + 33)

= 6 x 41

= 246

**S**

_{12}= 246_{ }

(iv)

To find: d, a_{10}

Given:

a_{3} = 15

S_{10} = 125

Sum of n terms can be given as follows:

S_{10}

125 = 5(2a + 9d)

25 = 2a + 9d .........................eq(i)

We know, nth term of an AP is

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

where a and d are first term and common difference respectively.

According to the question; the 3rd term is 15, which means;

a + 2d = 15 ...........eq(ii)

Now,

Subtracting equation 2 times eq(ii) from equation eq(i), we get;

(2a + 9d) – 2(a + 2d) = 25 – 30

Or, 2a + 9d - 2a - 4d = - 5

Or, 5d = - 5**d = -1**

Now,

Substituting the value of d in equation (2), we get;

a + 2(- 1) = 15

Or, a – 2 = 15

Or, a = 17

10^{th} term can be calculated as follows;

a_{10} = a + 9d

= 17 – 9 = 8

**a**

_{10}= 8Thus, d = - 1 and 10th term = 8

(v)**To find: a and a _{9}**

**Given:**

**d = 5**

**S**

_{9}= 75Sum of n terms can be given as follows:

75

75= (2a + 40)

= 2a x 40

Now,

We know, nth term of an AP is

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

where a and d are first term and common difference respectively.

9^{th} term can be calculated as follows:

a_{9} = a + 8d

= -

=

(vi) **To find : n and a _{n}**

**Given:**

**a = 2**

**d = 8**

**S**

_{n}= 90Sum of n terms can be given as follows:

90

90= (4 + 8N - 8)

90 = N(2 + 4N - 4)

4N^{2} - 2N - 90 =0

2N^{2} - N - 45 =0

(2N +9)(N -5)

Hence, n = - 9/2 and n = 5

Rejecting the negative value,

We have **n = 5**

We know, nth term of an AP is

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

where a and d are first term and common difference respectively.

Now, 5^{th }term will be:

_{5}= a + (5 - 1) d

a_{5} = a + 4d

= 2 + 4 x 8

= 2 + 32 = 34

**a**

_{5}= 34(vii) **To find: n and d****Given: ****a = 8****a _{n} = 62**

**S**

_{n}= 210Sum of n terms can be given as follows:

210

Since, a_{n} = a + (n - 1)d, we have

420 = (8 + 62)

420 = n x 70

**n = 6**

Now, for calculating d:

we know that a_{n}= a + (n - 1) d

a_{6} = a + 5d

Or, 62 = 8 + 5d

Or, 5d = 62 – 8 = 54

Or, d = 54/5

Common difference, d = 54/5

(viii) To find: n and a

Given:

a_{n} = 4

d = 2

S_{n} = - 14

Sum of n terms can be given as follows:

-14

-28 = (a + 4)

.................................eq(i)

We know;

We know, nth term of an AP is

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

where a and d are the first term and common difference respectively.

therefore,4 = a + (n – 1)2

Or, 4 = a + 2n – 2

Or, a + 2n = 6

Or, 2n = 6 – a

Or, n = (6 – a)/2 .............................eq(ii)

Using (i) and (ii):

Cross multiplying, we get

-56 = (6 – a)(a + 4)

24 + 2a – a^{2} = - 56

a^{2} - 2a - 80 = 0

(a + 8)(a – 10)= 0

Therefore, a = - 8 and a = 10

As a is smaller than 10 and d has positive value, hence we’ll take a = - 8

Now, we can find the number of terms as follows:

a_{n} = a + (n – 1)d

4 = - 8 + (n – 1)2

(n – 1)2 = 4 + 8 = 12

n – 1 = 6, n = 7

Hence, **n = 7 and a = - 8**

(ix) To find: d

Given: a =3

n = 8

S_{n} = 192

Sum of n terms can be given as follows:

192 = 4(6 + 7d)

7d = 42

**d = 6**

(x)**To Find: a****Given: l = 28, S = 144**

Sum of n terms can be given as follows:

288 = n (a + 28)

9a + 252 = 288

9a = 288 -252

9a = 36

**a = 4**

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The sum of n terms of an A.P. is 3n^{2}+ 5n. Find the A.P. Hence, find its 16th term.