Question

In an ap s10=136and s9=120then findt10​

Answer 1

$\sf\huge\bold{\underline{\underline{{Solution}}}}$

Given:

$\sf S_{10}= 136$

$\sf S_{9} = 120$

To Find

$\sf\huge t_{10}??$

We use sum formula because here sn is given

$\sf \bold{Sn = \dfrac{n}{2} \bigg[2a + \bigg(n - 1\bigg)d\bigg]}$

$\sf S_{10}= 136$

$\sf \longmapsto136 = \dfrac{10}{2} (2a + (10 - 1)d$

$\sf \longmapsto136 = 5(2a + 9d)$

$\sf \longmapsto10 + 45d = 136 - (1)$

$\sf S_{9} = 120$

$\sf \longmapsto120 = \dfrac{n}{2} (2a + (9 - 1)d$

$\sf \longmapsto120 = \dfrac{9}{2} (2a + 8d)$

$\sf \longmapsto120 = \dfrac{9}{2} \times 2(a + 4d)$

$\sf \longmapsto9a + 36d = 120 - (2)$

Now solving equation (1) and (2) by elimination method:

Now we will multiply with coefficient of 'a' to both the Equation :

$\sf \longmapsto10a + 45d = 136 (\times 9)$

$\sf \longmapsto9a + 36d = 120 (\times 10)$

New Equation obtained:

$\sf \longmapsto90a + 405d = 1224 - (3)$

$\sf \longmapsto90a + 360d = 1200 - (4)$

Now substracting Equation 4 from 3

=>90a+405d=1224

=>90a+360d=1200

=>(-)$\:\:$ (-)$\:\:\:\:\:\:\:$ (-)

_______________

45d=24

d=24/45=8/15

$\sf \bold{d = \dfrac{8}{15} }$

Now put d in Equation 1

$\sf \longmapsto10a + 45d = 136$

$\sf \longmapsto10a + 45 \times \dfrac{8}{15} = 136$

$\sf \longmapsto10a + 24 = 136$

$\sf \longmapsto10a = 112$

$\sf \longmapsto \: a = \dfrac{112}{10}=\dfrac{56}{5}$

$\sf \bold{a = \dfrac{56}{5} }$

We have find a and d now we have to find $\sf t_{10}$

$\sf \longmapsto \: t10 = a + (n - 1)d$

$\sf \longmapsto \dfrac{56}{5} + (10 - 1) \times \dfrac{8}{15}$

$\sf \longmapsto \dfrac{56}{5} + 9 \times \dfrac{8}{15}$

$\sf \longmapsto \dfrac{56}{5} + 3 \times \dfrac{8}{5}$

$\sf \longmapsto \dfrac{56}{5} + \dfrac{24}{5} = \dfrac{80}{5} = 16$

$\sf\bold{ \therefore \: t_{10} = 16}$