Question

Find ${20^{th}}$ term from the end of an A.P. $3,7,11,........407.$

Answer 1

Hint: In an A.P. \[{m^{th}}\] term from the end is equal to ${(n - m + 1)^{th}}$ term from the beginning i.e., ${a_m} = {a_{(n - m + 1)}}$
Hence first find the number of terms, then use the above formula.

Complete step by step answer:
In an AP, find first term $(a) = 3$
Common difference $(d) = {a_2} - {a_1}$$ = 7 - 3 = 4$
Last term $({a_n}) = 407$
Using formula of ${n^{th}}$ term, find number of terms
${a_n} = a + (n - 1)d$
$407 = 3 + (n - 1)4$
$404 = (n - 1)4$
$n - 1 = 101$
$n = 102$
Now ${a_m}$ (from the end) $ = {a_{(n - m + 1)}}$ (from the beginning)
${a_{20}} = {a_{(102 - 20 + 1)}}$
${a_{20}} = {a_{83}}$
${a_{83}} = a + (83 - 1)d$
$ = 3 + 82 \times 4$
$ = 331$

Note: Alternate solution –
In an AP ${m^{th}}$ term from the end can be calculated by taking AP in reverse order.
${a_m} = {a_n} + (m - 1)( - d)$
In reverse order –
First term will be ${a_n}$
Common difference will be negative of difference in forward order
${a_m} = {a_n} + (m - 1)( - d)$
${a_{20}} = 407 + (20 - 1)( - 4)$
$ = 407 - 76$
$ = 331$