Class 10

Math

All topics

Sequences and Series

For an A.P., the seventh term is 19 and the thirteenth term is 37. Find its twenty second term.

- 60
- 54
- 64
- 78

Now $T_{n}=a+(n−1)d$

$∴T_{7}=a+(7−1)d$

$∴19=a+6d$

$∴a+6d=19$ ............ (1)

Again, $T_{13}=a+(13−1)d$

$∴37=a+12d$

$∴a+12d=37$ ............. (2)

Subtracting eq. (1) from eq. (2)

$a +12d =37$

$a +6d =19$

$6d=18$

$∴d=13$

Substituting $d=13$ in $a+6d=19$,

$a+6(3)=19$

$∴a+18=19$

$∴a=1$

Thus, the first term of the A.P. is 1 and the common difference is 3.

Now, $T_{22}=a+(22−1)d=1+(22−1)(3)$

$=1+(21)(3)=1+63=64$

Thus, the twenty second term of the given A.P. is 64.

Substituting $d=13$ in $a+6d=19$,

$a+6(3)=19$

$∴a+18=19$

$∴a=1$

Thus, the first term of the A.P. is 1 and the common difference is 3.

Now, $T_{22}=a+(22−1)d=1+(22−1)(3)$

$=1+(21)(3)=1+63=64$

Thus, the twenty second term of the given A.P. is 64.