Question

7th term of arithmetic sequence is32.find first and 13th terms

Answer 1

Answer:

Given, 7th term of A.P

$a7=32$

and 13th term of A.P is

$a13=62$

$nth \: term \: of A.P \: is \: given \: by</p><p></p><p>an=a1+(n−1)d</p><p></p><p>hence 7th term is given by</p><p></p><p>a7=a1+(7−1)d</p><p></p><p>put a7=32 which is given</p><p></p><p>32=a1+6d...….eq(1)</p><p></p><p>13th \: term \: is \: given \: by</p><p></p><p>a13=a1+(13−1)d</p><p></p><p>put  \: a13=62 which \: is \: given</p><p></p><p>62=a1+12d...….eq(2)</p><p></p><p>substract \: eq(1)  from \: eq(2) we \: get</p><p></p><p>62−32=a1−a1+12d−6d</p><p></p><p>⟹30=6d</p><p></p><p>⟹d=630</p><p></p><p>⟹d=5.....eq(3)  \: common \: difference</p><p></p><p>put  \: d=5 in eq(1) we get</p><p></p><p>32=a1+6×5</p><p></p><p>⟹a1=32−30</p><p></p><p>⟹a1=2 first \: term \: of \: A.P</p><p></p><p>2nd \: term \: of \: A.Pa2=a1+d=2+5=7</p><p></p><p>3rd \:  term \: of \: A.Pa3=a2+d=7+5=12</p><p></p><p>4th  \: term of \: A.Pa4=a1+d=12+5=17</p><p></p><p>and \: so \: on</p><p></p><p>hence \: the \: A.P is </p><p></p><p>2,7,12,17,...…..</p><p></p><p>$

Step-by-step explanation:

hope it helped you friend if it helps you mark it as brainliest