Question

34. Find the sum of first 50 terms of an AP,
whose second and third terms are 14 and
18 respectively.​

Answer 1

Answer:

✯ ANSWER REFER TO THE ATTACHMENT✯

(◕ᴗ◕✿)

Answer 2

$\mathfrak{{\pmb{{\underline{Given}}:}}}$

• The second and third terms are 14 and 18 respectively (i.e., $\sf{a_{2}=14~,~~a_{3}=18}$)

$\mathfrak{{\pmb{{\underline{To~find}}:}}}$

• The sum of first 50 terms of the AP

$\mathfrak{{\pmb{{\underline{Solution}}:}}}$

$\sf{{\pmb{{\underline{Given}}:}}}$

• $\sf{a_{2}=14}$
• $\sf{a_{3}=18}$

$\sf{~~~~ \therefore d=a_{3}-a_{2}=18-14={\pmb{4}}}$

$\sf{~~~~~~~~~~ a_{2}=a+d}$

$\sf\implies{14=a+4}$

$\sf\implies{a=14-4}$

$\sf\implies{ {\blue{\underline{\boxed{\sf{\pmb{a=10}}}}}}}$

$\sf{~~~~ \therefore a={\pmb{10}}}$

• Sum of first 50 terms

$\sf{~~~~ \therefore n={\pmb{50}}}$

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

$\sf\implies{S_{50}={\dfrac{50}{2}}[2(10)+(50-1)4]}$

$\sf\implies{S_{50}=25[20+(49)4]}$

$\sf\implies{S_{50}=25(20+196)}$

$\sf\implies{S_{50}=25(216)}$

$\sf\implies{ {\blue{\underline{\boxed{\sf{\pmb{S_{50}=5,400}}}}}}}$

$\therefore\mathfrak{{\pmb{{\underline{Required~answer}}:}}}$

Sum of first 50 terms of an AP is 5,400