Question

For an AP if s10 = 150 and s9 = 126, t10 = ?

Answer 1

Answer:

$t_{10} = 24$

Step-by-step explanation:

Given:-

$s_{10} = 150 \\ s_{9} = 126 \\ t_{10} = \\ \\ now \: as \: we \: know \\ \\ s_{n} = \frac{n}{2} (2a + (n - 1)d) \\ \\ s_{10} = \frac{10}{2} (2a + (10 - 1)d) \\ \\ 150 = 5(2a + 9d) \\ \\ \frac{150}{5} = (2a + 9d) \\ \\ (2a + 9d) = 30\: \: \: .................(1) \\ \\ \\ s_{n} = \frac{n}{2} (2a + (n - 1)d) \\ \\ s_{9} = \frac{9}{2} (2a + (9 -1 )d \\ \\ 126 = \frac{9}{2} (2a + 8d) \\ \\ \frac{126 \times 2}{9} = (2a + 8d) \\ \\ (2a + 8d) = 28 \: \: ...............(2) \\ \\ now \: substract \: equation \: 1 \: \: and \: 2 \\ \\ \: \: \: 2a + 9d = 30 \\ - \: 2a + 8d = 28 \\ - - - - - - - \\ d = 2 \\ \\ now \: put \: d \: = 2 \: in \: equation \: \: 1 \\ \\ 2a + 9d = 30 \\ \\ 2a + 9 \times 2 = 30 \\ 2a = 30 - 18 \\ \\ 2a = 12 \\ \\ a = \frac{12}{2} \\ a = 6 \\ \\ \\ now \: we \: have \: a = 6 \: and \: d = 2 \\ \\ there \: for \\ \\ as \: we \: know \\ \\ t_{n} = a + (n - 1)d \\ \\ t_{10} = 6 + (10 - 1) \times 2 \\ t_{10} = 6 + 9 \times 2 \\ t_{10} = 6 + 18 \\ \\ t_{10} = 24$

Answer 2

$t_{10} =24$

Step-by-step explanation:

Let the first term = a and common difference = d

Given,

$s_{10} = 150$ and $s_{9} = 126$

To find, the value of $t_{10}= ?$

We know that,

The sum of nth term of an AP

$s_{n} =\dfrac{n}{2}{2a+(n-1)d}$

⇒ $s_{10} =\dfrac{10}{2}[2a+(10-1)d]$

⇒ $5[2a+9d]=150$

⇒ $2a+9d=30$             ....... (1)

Also, $s_{9}=\dfrac{9}{2} [2a+8d]$

⇒ $a+4d=\dfrac{126}{9} =14$     ....... (2)    

Multiplying (2) by 2 and and subtracting (1), we get

9d - 8d = 30 - 28 = 2

⇒ d = 2

Put d = 2 in equation (1), we get

$2a+9(2)=30$

⇒ 2a = 30 - 18

⇒ a = 6

∴ The 10th term of an AP

$t_{10} =a+9d$

= 6 + 9 (2)

= 6 + 18 = 24

Hence, $t_{10} =24$