Question

the first and the fourth term of an ap are 12 and 20 respectively find the 25 term of the AP​

Answer 1

Answer:

★ 25th term = 76 ★

Step-by-step explanation:

Given:

• The first and the fourth term of an AP are 12 and 20 respectively.

To find:

• 25th term of the AP.

Solution:

We know that,

${\boxed{\sf{T_n=a+(n-1)d}}}$

Here,

• a = first term
• d = common difference

Now find the first and 4th term of the AP.

$\implies\sf{T_1=a+(1-1)d}$

$\implies\sf{12=a.......(i)}$

&

$\implies\sf{T_4=a+(4-1)d}$

$\implies\sf{T_4=a+3d}$

• Now put a=12 from eq(i).

$\implies\sf{20=12+3d}$

$\implies\sf{3d=20-12}$

$\implies\sf{d=8/3}$

Now find the 25th term of the AP.

$\mapsto\sf{T_{25}=12+(25-1)\times\dfrac{8}{3}}$

$\mapsto\sf{T_{25}=12+24\times\dfrac{8}{3}}$

$\mapsto\sf{T_{25}=12+64}$

$\mapsto\sf{T_{25}=76}$

Hence, the 25th term of the AP is 76.

Answer 2

Given:-

• $\sf 1^{st}$ term of AP = 12
• $\sf 4^{th}$ term of AP = 20

Find:-

• $\sf 25^{th}$ term of AP

Solution:-

we, know that

$\huge{\underline{\boxed{\sf a_n = a+(n-1)d}}}$

$\pink{\sf where } \footnotesize{\begin{cases} \purple{ \sf a = 1} \\ \red{\sf n = 1 }\\ \orange{\sf a_n = 12} \end{cases}}$

☯$\footnotesize{\underline{Substituting\:these\:values\:in\:the\: formula:-}}$

$\implies\sf a_n = a+(n-1)d$

$\implies\sf 12 = a+(1-1)d$

$\implies\sf 12 = a+(0)d$

$\implies\sf 12 = a+0$

$\implies\sf 12 =a \qquad \bigg \lgroup Equation \: 1 \bigg \rgroup$

$\therefore\sf a = 12$

Again, using

$\huge{\underline{\boxed{\sf a_n = a+(n-1)d}}}$

$\green{\sf where } \footnotesize{\begin{cases} \pink{ \sf a = 12} \qquad \lgroup \sf Equation\:1 \rgroup \\ \purple{\sf n = 4}\\ \pink{\sf a_n = 20} \end{cases}}$

☯$\footnotesize{\underline{Substituting\:these\:values\:in\:the\: formula:-}}$

$\dashrightarrow\sf a_n = a+(n-1)d$

$\dashrightarrow\sf 20 = 12+(4-1)d$

$\dashrightarrow\sf 20 - 12=(3)d$

$\dashrightarrow\sf 8=3d$

$\dashrightarrow\sf \dfrac{8}{3}=d$

$\therefore\sf d = \dfrac{8}{3}$

Using,

$\huge{\underline{\boxed{\sf a_n = a+(n-1)d}}}$

$\pink{\sf where } \footnotesize{\begin{cases} \green{ \sf a = 12} \qquad \lgroup \sf Equation\:1 \rgroup \\ \blue{\sf n = 25}\\ \purple{\sf d = \dfrac{8}{3} } \end{cases}}$

☯$\footnotesize{\underline{Substituting\:these\:values\:in\:the\: formula:-}}$

$\dashrightarrow\sf a_{25} = a+(n-1)d$

$\dashrightarrow\sf a_{25} = 12+(25-1) \times \dfrac{8}{3}$

$\dashrightarrow\sf a_{25}= 12+(24) \times \dfrac{8}{3}$

$\dashrightarrow\sf a_{25} = 12+24\times \dfrac{8}{3}$

$\dashrightarrow\sf a_{25} = 12 + \dfrac{192}{3}$

$\dashrightarrow\sf a_{25} = 12+64$

$\dashrightarrow\sf a_{25} = 76$

$\small{\therefore \underline{\sf 25^{th}\:term\:of\:the\:A.P. \: is \: 76}}$