Question

find 20th term from the last term of the ap 2,7,12.........212
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Answer 1

Answer:

$\sf{The \ 20^{th} \ term \ from \ the \ last \ is \ 117.}$

Given:

• The given A.P. is 2, 7, 12,...,212

To find:

• $\sf{20^{th} \ term \ from \ the \ last.}$

Solution:

$\sf{The \ given \ A.P. \ is}$

$\sf{2, \ 7, \ 12,...,212}$

$\sf{Here, \ first \ term(a)=2}$

$\sf{Common \ difference (d)=7-2=5}$

$\sf{n^{th} \ term(t_{n})=212}$

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

$\sf{\leadsto{212=2+(n-1)\times5}}$

$\sf{\leadsto{\therefore{5(n-1)=212-2}}}$

$\sf{\leadsto{\therefore{5(n-1)=210}}}$

$\sf{\leadsto{\therefore{n-1=\dfrac{210}{5}}}}$

$\sf{\leadsto{\therefore{n-1=42}}}$

$\sf{\leadsto{\therefore{n=42+1}}}$

$\sf{\leadsto{\therefore{n=43}}}$

$\sf{2^{nd} \ term \ from \ last=t_{n-1}}$

$\sf{\therefore{20^{th} \ term \ from \ last=t_{n-19}}}$

$\sf{\leadsto{t_{n-19}=a+(n-19-1)d}}$

$\sf{\leadsto{\therefore{t_{n-19}=2+(43-20)\times5}}}$

$\sf{\leadsto{\therefore{t_{n-19}=2+(23\times5)}}}$

$\sf{\leadsto{\therefore{t_{n-19}=2+115}}}$

$\sf{\leadsto{\therefore{t_{n-19}=117}}}$

$\sf\purple{\tt{\therefore{The \ 20^{th} \ term \ from \ the \ last \ is \ 117.}}}$