Question

which term of the Ap 5,15,25,....will be 140 more than 31st term​

Answer 1

Answer:

45th term

Step-by-step explanation:

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Answer 2

${\huge{\boxed{\underline{\textsf{\textbf{\color{navy}{Final\:An}{\pink{sw}{\purple{er:-}}}}}}}}}$

$\boxed{ 45^{th} \: term}$

$\huge\underbrace\red{\dag Given:- \dag}$

Ap 5,15,25,..

$\huge\underbrace\pink {\dag To Find:- \dag}$

to find the term which is 140 more than 31st term

$\huge\underbrace\green {\dag Formula\:Used:- \dag}$

• $an = a+(n-1)d$

• $d=a2-a1$

$\huge\underbrace\purple{\dag Solution:- \dag}$

Ap is $5,15,25..$

So let’s find the $31st \ term$

$a31 = a + (n-1)d\\a31 = 5 + (31-1)10\\a31 = 5 +(30)10\\a31 = 5+300\\a31 = 305$

now the term more than 140 of $31st\: term \implies 305+140 \implies 445$

therefore let’s find the term of 445  

$a_n = a+(n-1)d\\445=5+(n-1)10 \\445=5+10n-10\\445=10n-5\\10n=445+5\\10n=450$

$n=\frac{450}{10}$

$n=45$

hence the term $31^{st}$  more than 140 $(445) \implies 45^{th} \: term$

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$\huge\underbrace\blue{\dag things \ to \ know :-\dag}$

an= term which we have to find

n= number of term in series or ap

d= common difference

a= first term of ap