Question

There are 37 terms in an A.P ., the sum of three terms placed exactly at the middle is 225 and the sum of last three terms is 429 .write the A.P. guys
plzz it's urgent....​

Answer 1

Step-by-step explanation:

Hope this will help u

like my answer

Answer 2

Solution :

$\underline{\bf{Given\::}}$

There are 37 terms in an A.P., the sum of three terms placed exactly at the middle is 225 & the sum of last three terms is 429 .

$\underline{\bf{Explanation\::}}$

As we know that formula of the middle term, we have odd terms = 37 so;

$\boxed{\bf{Middle\:terms = \bigg(\frac{n+1}{2} \bigg)^{th}}}$

$\mapsto\tt{Middle\:term =\bigg( \dfrac{37+1}{2} \bigg)^{th}}$

$\mapsto\tt{Middle\:term =\bigg( \cancel{\dfrac{38}{2} }\bigg)^{th}}$

$\mapsto\tt{Middle\:term = 19^{th}\:term }$

As we know that formula of an A.P;

$\boxed{\bf{a_n=a+(n-1)d}}$

• a is the first term .
• d is the common difference .
• n is the term of an A.P.

$\underline{\underline{\tt{According\:to\:the\:question\::}}}$

$\underbrace{\bf{1^{st}\:condition\::}}$

$\longrightarrow\tt{a_{18} + a_{19} + a_{20} = 225}$

$\longrightarrow\tt{a+(18-1)d + a+(19-1)d + a+(20-1)d = 225}$

$\longrightarrow\tt{a+17d + a+18d + a+19d= 225}$

$\longrightarrow\tt{3a+54d= 225}$

$\longrightarrow\tt{3(a+18d)= 225}$

$\longrightarrow\tt{a+18d= \cancel{225/3}}$

$\longrightarrow\tt{a+18d= 75}$

$\longrightarrow\tt{a=75-18d..................(1)}$

$\underbrace{\bf{2^{nd}\:condition\::}}$

$\longrightarrow\tt{a_{35} + a_{36} + a_{37} = 429}$

$\longrightarrow\tt{a+(35-1)d + a+(36-1)d + a+(37-1)d = 429}$

$\longrightarrow\tt{a+34d + a+35d + a+36d= 429}$

$\longrightarrow\tt{3a+105d= 429}$

$\longrightarrow\tt{3(a+35d)= 429}$

$\longrightarrow\tt{a+35d= \cancel{429/3}}$

$\longrightarrow\tt{a+35d= 143}$

$\longrightarrow\tt{75-18d+35d= 143\:\:\:[from(1)]}$

$\longrightarrow\tt{75+17d= 143}$

$\longrightarrow\tt{17d= 143-75}$

$\longrightarrow\tt{17d= 68}$

$\longrightarrow\tt{d= \cancel{68/17}}$

$\longrightarrow\bf{d=4}$

∴ Putting the value of d in equation (1),we get;

$\longrightarrow\tt{a=75-18(4)}$

$\longrightarrow\tt{a=75-72}$

$\longrightarrow\bf{a=3}$

Thus;

A R I T H M E T I C   P R O G R E S S I O N :

$\bullet\sf{a=\boxed{\bf{3}}}$

$\bullet\sf{a+d=3+4=\boxed{\bf{7}}}$

$\bullet\sf{a+2d=3+2(4) = 3+8=\boxed{\bf{11}}}$

$\bullet\sf{a+3d=3+3(4) = 3+12=\boxed{\bf{15}}}$