Question

The sum of first six terms of an A.P is 42. The ratio of 10th term to the 30th term is 1 : 3. Find the 13th term.​

Answer 1

Answer:

Sum of the first n terms is given by

S

n

=

2

n

[2a+(n−1)d]

S

6

=

2

6

[2a+(6−1)d]

S

6

=3[2a+5d]

T

30

T

10

=

3

1

a+29d

a+9d

=

3

1

(a+9d)3=(a+29d)1

3a+27d=a+29d

2a=2d

a=d...(1)

S

6

=3[2a+5d]

S

6

=3[2a+5a]

42=3(7a)

42=21a

∴a=2

From (1)

d=a=2

T

13

=a+(n−1)d

=2+(13−1)2

=2+24

=26

Answer 2

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Sum of the first n terms is given by

S

n

=

2

n

[2a+(n−1)d]

S

6

=

2

6

[2a+(6−1)d]

S

6

=3[2a+5d]

T

30

T

10

=

3

1

a+29d

a+9d

=

3

1

(a+9d)3=(a+29d)1

3a+27d=a+29d

2a=2d

a=d...(1)

S

6

=3[2a+5d]

S

6

=3[2a+5a]

42=3(7a)

42=21a

∴a=2

From (1)

d=a=2

T

13

=a+(n−1)d

=2+(13−1)2

=2+24

=26