Question

for an A. P. if s10 =150 &s9 =126, t10? ​

Answer 1

Answer:

$\large\bold\red{{ t}_{ 10}=24}$

Step-by-step explanation:

Given,

In an A.P,

${ s}_{ 10} = 150$

And,

${ s}_{9 } = 126$

But,

We kniw that,

$\large \boxed{ \bold{{ s}_{n } = \frac{n}{2} (a + l)}}$

Where,

• a = 1st term
• l = Last term

Therefore,

We get,

$= > \dfrac{10}{2} (a + { t}_{ 10}) = 150 \\ \\ = > a + { t}_{10 } = \dfrac{150}{5} \\ \\ = > a + { t}_{10 } = 30 \: \: \: \: .......(1)$

And,

$= > \frac{9}{2} (a + { t}_{9 }) = 126 \\ \\ = > a + { t}_{ 9} = \dfrac{126}{9} \times 2 \\ \\ = > a + { t}_{9 } = 14 \times 2 \\ \\ = > a + { t}_{ 9} = 28 \: \: \: \: .......(2)$

Subtracting eqn (2) from (1),

We get,

$= > { t}_{ 10} - { t}_{ 9} = 2$

Let,

• Common difference = d

Therefore,

We get,

$= > d = 2$

Therefore,

We get,

$= > \frac{10}{2} (2a + 2(10 - 1)) = 150 \\ \\ = > 2a + 18 = 30 \\ \\ = > 2a = 30 - 18 = 12 \\ \\ = > a = \dfrac{12}{2} \\ \\ = > a = 6$

Thus,

We have,

${ t}_{ 10} = a + 9d \\ \\ = > { t}_{ 10} = 6 + (9 \times 2 )= 6 + 18 \\ \\ = > \bold{ { t}_{ 10} = 24}$