Question

5th term of a sequence is 50 and the 10th term is 70, then

(a) find the common difference of this sequence.

(b) write down the sequence.

(c) calculate the sum of first 20 terms of this sequence.​

Answer 1

Given :-

5th term ($T_{5}$)= 50

10th term ($T_{10}$) = 70

Formula to be used :-

$T_{n} = a + (n-1)d$

$Sum\:\:of\:\:n\:\:terms\:\:(S_{n}) = \frac{n}{2} (2a+(n-1)d)$

Here,

a = first term

d = common difference

n = number of terms

Solution :-

a.) 5th term ($T_{5}$) = $a + (5-1)d$

=> 50 = a + 4d                   [5th term ($T_{5}$)= 50]

=> 50 - 4d = a                    ....(i.)

=> 10th term ($T_{10}$) = $a + (10-1)d$

=> 70 = a + 9d

Substituting the value of a from eq. (i.)

=> 70 = 50 - 4d + 9d  

=> 70 = 50 + 5d

=> 70 - 50 = 5d

=> 20 = 5d

=> $\frac{20}{5}$ = d

=> 4 = d

Now,

=> a = 50 - 4d

=> a = 50 - 4 x 4

=> a = 50 - 16

=> a = 34

b.) 1st term =  $a + (1-1)d$  = a + 0 = 34 + 0 = 34

2nd term =  $a + (2-1)d$  = a + d = 34 + 4 = 38

3rd term =  $a + (3-1)d$  = a + 2d = 34 + 2 x 4 = 34 + 8 = 42

4th term =  $a + (4-1)d$  = a + 3d = 34 + 3 x 4 = 34 + 12 = 46

5th term =  $a + (5-1)d$  = a + 4d = 34 + 4 x 4 = 34 + 16 = 50

6th term =  $a + (6-1)d$  = a + 5d = 34 + 5 x 4 = 34 + 20 = 54

7th term =  $a + (7-1)d$  = a + 6d = 34 + 6 x 4 = 34 + 24 = 58

8th term =  $a + (8-1)d$  = a + 7d = 34 + 7 x 4 = 34 + 28 = 62

9th term =  $a + (9-1)d$  = a + 8d = 34 + 8 x 4 = 34 + 32 = 66

10th term =  $a + (10-1)d$  = a + 9d = 34 + 9 x 4 = 34 + 34 = 70

Sequence:

34, 38, 42, 46, 50, 54, 58, 62, 66, 70.

c.) $Sum\:\:of\:\:n\:\:terms\:\:(S_{n}) = \frac{n}{2} (2a+(n-1)d)$

$S_{20}= \frac{20}{2} (2 \times 34 +(20-1)4) \\ \\ = 10(2 \times 34 +19\times 4)\\ \\ = 10(68 +76)\\\\=10\times 144\\\\1440$

Answer:-

a.) The common difference = 4

b.) Sequence: 34, 38, 42, 46, 50, 54, 58, 62, 66, 70.

c.) The sum of first 20 terms of this sequence = 1440

Answer 2

Answer:

a.) The common difference = 4

b.) Sequence: 34, 38, 42, 46, 50, 54, 58, 62, 66, 70.

c.) The sum of first 20 terms of this sequence = 1440

Step-by-step explanation:

Given :-

• 5th term ($T_{5}$)= 50
• 10th term ($T_{10}$) = 70

Formula to be used :-

$T_{n} = a + (n-1)d$

$Sum\:\:of\:\:n\:\:terms\:\:(S_{n}) = \frac{n}{2} (2a+(n-1)d)$

Here,

a = first term

d = common difference

n = number of terms

Solution :-

a.) 5th term ($T_{5}$) = $a + (5-1)d$

=> 50 = a + 4d                   [5th term ($T_{5}$)= 50]

=> 50 - 4d = a                    ....(i.)

=> 10th term ($T_{10}$) = $a + (10-1)d$

=> 70 = a + 9d

Substituting the value of a from eq. (i.)

=> 70 = 50 - 4d + 9d  

=> 70 = 50 + 5d

=> 70 - 50 = 5d

=> 20 = 5d

=> $\frac{20}{5}$ = d

=> 4 = d

Now,

=> a = 50 - 4d

=> a = 50 - 4 x 4

=> a = 50 - 16

=> a = 34

b.) 1st term =  $a + (1-1)d$  = a + 0 = 34 + 0 = 34

2nd term =  $a + (2-1)d$  = a + d = 34 + 4 = 38

3rd term =  $a + (3-1)d$  = a + 2d = 34 + 2 x 4 = 34 + 8 = 42

4th term =  $a + (4-1)d$  = a + 3d = 34 + 3 x 4 = 34 + 12 = 46

5th term =  $a + (5-1)d$  = a + 4d = 34 + 4 x 4 = 34 + 16 = 50

6th term =  $a + (6-1)d$  = a + 5d = 34 + 5 x 4 = 34 + 20 = 54

7th term =  $a + (7-1)d$  = a + 6d = 34 + 6 x 4 = 34 + 24 = 58

8th term =  $a + (8-1)d$  = a + 7d = 34 + 7 x 4 = 34 + 28 = 62

9th term =  $a + (9-1)d$  = a + 8d = 34 + 8 x 4 = 34 + 32 = 66

10th term =  $a + (10-1)d$  = a + 9d = 34 + 9 x 4 = 34 + 34 = 70

Sequence:

34, 38, 42, 46, 50, 54, 58, 62, 66, 70.

c.) $Sum\:\:of\:\:n\:\:terms\:\:(S_{n}) = \frac{n}{2} (2a+(n-1)d)$

$S_{20}= \frac{20}{2} (2 \times 34 +(20-1)4) \\ \\ = 10(2 \times 34 +19\times 4)\\ \\ = 10(68 +76)\\\\=10\times 144\\\\1440$