Question

The 5th term of an ap is 24 and its 15th term is 74. find the sum of its first 10 terms

Answer 1

T5=24
T15=74
S10=?

a+4d=24
a+14d=74
(-) (-) (-)
-10d=-50
d=50/10=5
a+4d=24
a+4 (5)=24
a+20=24
a=24-20
a=4

a=4 d=5 n=10
Sn=n/2 [2a+(n-1)d]
S10=10/2 [2 (4)+(10-1)5]
S10=5 (8+45)
S10=5 (53)
S10=265

Therefore sum of 10 terms is 265.

Answer 2

The sum of its first $10^{th}$ terms is $265$

Step-by-step explanation:

Given,

The $5^{th}$ term of an AP is $24$ and its $15^{th}$  term is $74$.

∴ $a_{5} =a+4d$

And given,

$a_{5} =24$

So,

$a+4d=24$ __1

Similarly,

$a_{15} =a+14d$

Given,

$a_{15} =74$  

So,

$a+14d=74$ __2

Subtract equation-1 & 2,

$a+4d-a-14d=24-74$

⇒$-10d=-50$

⇒$d=5$

Plug $d$ value in equation-1,

$a=24-(4\times 5)$

⇒$a=4$

Sum of first $n^{th}$ terms is given,

$S_{n}=\frac{n}{2}[2a+(n-1)d]$

Here, $n=10$,

∴ $S_{10}=\frac{10}{2}[(2\times 4)+(10-1)5]$

⇒$S_{10}=5[8+45]$

⇒$S_{10}=265$

So, The sum of its first $10^{th}$ terms is $265$