Question

For an a.p.if t3=8 and t7=24,then t10 =

Answer 1

$\bold{For \: an \: AP \: }$

$\huge\boxed{t_n = a + (n-1)d}$

Let the first term be a

Common difference be d

Given,

$\mathsf{ t_7= 24}$

$\mathsf{t_7= a + (7-1)d}$

$\mathsf{24= a + 6d}$------(1)

$\mathsf{t_3= 8}$

$\mathsf{ t_3= a + (3-1)d}$

$\mathsf{8= a + 2d}$------(2)

On solving the eq 1.and 2

simultaneously,

$\mathsf{8= a + 2d}$

$\mathsf{ 24= a + 6d}$

______________________

-16 = -4d

________________________

$\mathsf{\dfrac{-16}{-4} = d}$

$\huge \boxed{\mathsf{4 = d}}$

On placing value of d in eq 2

$\mathsf{8= a + 2\times 4}$

$\mathsf{8 - 8= a}$

$\huge \boxed{\mathsf{0= a}}$

For tenth term,

$\mathsf{t_10= ?}$

$\mathsf{t_10= 0 + (10 -1) \times 4}$

$\mathsf{t_10= 0+ 9 \times 4}$

$\huge \boxed{\mathsf{t_{10}= 36}}$