Question

Sum of the first 14 terms of an A.P. is 1505 and its first term is 10 . Find its 25^th term

Answer 1

GivEn:-

• First term of AP (a) = 10

• Sum of the first 14 terms of an A.P. is 1505

To find:-

• $\sf 25^{th}$ term of AP ($\sf a_{25}$

SoluTion:-

As we know that,

✩ Sum of n terms of an AP -

$\dag\;{\underline{\boxed{\bf{\pink{S_n = \dfrac{n}{2} \bigg( 2a + (n - 1)d \bigg)}}}}}$

$\;\;\;\;\star\;\sf \underline{Put\;givEn\;values:-}$

$:\implies\sf 1505 = \dfrac{14}{2} \bigg( 2 \times 10 + (14 - 1)d \bigg)$

$:\implies\sf 1505 \times 2 = 14( 20 + 13d)$

$:\implies\sf 3010 = 14( 20 + 13d)$

$:\implies\sf \cancel{ \dfrac{3010}{14}} = 20 + 13d$

$:\implies\sf 215 = 20 + 13d$

$:\implies\sf 215 - 20 = 13d$

$:\implies\sf 195 = 13d$

$:\implies\sf d = \cancel{ \dfrac{195}{13}}$

$:\implies{\underline{\boxed{\bf{\purple{ d = 15}}}}}$

▬▬▬▬▬▬▬▬▬▬

As we know that,

✩ $\sf n_{th}$ terms of an AP -

$\dag\;{\underline{\boxed{\bf{\pink{a_n = 2a + (n - 1)d}}}}}$

$\;\;\;\;\star\;\sf \underline{Put\;givEn\;values:-}$

$:\implies\sf a_{25} = 10 + (25 - 1) 15$

$:\implies\sf a_{25} = 10 + 24 \times 15$

$:\implies\sf a_{25} = 10 + 360$

$:\implies{\underline{\boxed{\bf{\purple{ a_{25} = 370}}}}}$

$\therefore\;\sf \underline{25^{th} \; term \; of \; AP \; is \; \bf{370}}$

▬▬▬▬▬▬▬▬▬▬

★ Formula Used:-

$\begin{lgathered}\boxed{\begin{minipage}{15 em} \sf \displaystyle \bullet a_n=a + (n-1)d \\\\\\ \bullet S_n= \dfrac{n}{2} \left(a + a_n\right) \end{minipage}}\end{lgathered}$

▬▬▬▬▬▬▬▬▬▬

Answer 2

Answer:

$\large\boxed{\star\:\:\: a_{25}=370\:\:\:\star}$

Step-by-step explanation:

Given :–

• S₁₄=1505
• a₁=a=10

To Find :–

• a₂₅ (25th term of this A.P.)

Formulas Applied :–

• $a_n=a+(n-1)d$

• $S_n=\frac{n}{2} [2a+(n-1)d]$

Solution :–

We are given that : S₁₄=1505, a=10, n=14.

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

Put the above values in this formula :-

$\implies S_{14}=\frac{14}{2}[2(10)+(14-1)d]$

$\implies 1505=\frac{14}{2}[2(10)+(14-1)d] \:\:\:\:\:\:\:\:\:[\because\:S_{14}=1505]$

$\implies 1505=7\times(20+13d)$

$\implies \frac{1505}{7} =20+13d$

$\implies 215=20+13d$

$\implies 13d=215-20$

$\implies 13d=195$

$\implies d=\frac{195}{13}$

$\implies \boxed{d=15}$

Now ,we have to find a₂₅ :

For finding a₂₅, we have :

• a=10
• d=15
• n=25

aₙ=a+(n-1)d

Put the above values in this formula :-

$\implies a_{25}=10+(25-1)\times(15)$

$\implies a_{25}=10+(24)\times(15)$

$\implies a_{25}=10+360$

$\implies \boxed{a_{25}=370}$

→ More Information about A.P. :–

A.P. (Arithmetic Progression) is a sequence of series which have a Common Difference(d).

Some more Formulas From A.P. :

The formulas below are used when the Last term of the A.P. is Given.

• $S_n=\frac{n}{2} [a+l] \:,\:\:where \:l\:denotes\:the\:last\:term.$

• $a_n=l-(n-1)d\:,\:where\:l\:denotes\:the\:Last\:Term.$