Question

In an A. P. first term is 2 and sum of first five terms is one-fourth of sum of next five terms. so show that 20th term is -112

Answer 1

Solution:

Given:The first term of an AP is 2,
which means,    $a=2$...(i)

The sum of the first 5 terms is one-fourth of the sum of next 5 terms,
which means,    $S_{5}=(S_{10}-S_{5})/4$ ...(ii)

To show: $a_{20}=-112$

Proof: By using $S_n$ formula,
           
            $S_n=n(a+l)/2$
            $S_{5}=(S_{10}-S_{5})/4$
            $5(2+a_{5})/2=[10(2+a_{10})/2-5(2+a_{5})/2]/4$
            $10+5a_5=[10(2+a_{10})-(10+5a_5)]/4$
            $4(10+5a_5)=10(2+a_{10})$
            $5(10+5a_5)=10(2+a_{10})$
            $50+25a_5=20+10a_{10}$
            $50-20=10a_{10}-25a_5$
            [tex]30=5(2a_{10}-5a_5) [/tex]
            $6=(2a_{10}-5a_5)$
            $6=2(a+9d)-5(a+4d)$
            [tex]6=2(2+9d)-5(2+4d) [/tex]
            $6=4+18d-10-20d$
            $6=-6-2d$
            $-2d=12$
            $d=-6$  ...(iii)
           
          $a_{20}<strong>=</strong><strong>a+19d</strong>$
                     $<strong>=2+19(-6)</strong>$
                     $<strong>=2-114</strong>$
                     $<strong>=-112</strong>$
            $<strong></strong><strong>∴[tex]a_{20}=-112$=[]