Question

Find the value of $T_n$

(i) $1+4+7+10$ to $n,n=590$.

(ii) $50+46+42+38$ to $n,n=336.$

Answer 1

(1).

We have,

$1+4+7+10+......$

$n=590$

$a=1$

$d=4-1=3$

We know that,

$T_n=1+(590-1)\times 3$

$=1+589\times 3$

$=1+1767$

$T_n=1768$

Hence, this is the answer.

(2):-

$50+46+42+38+.......$

$n=336$

$a=50$

$d=46-50$

$d=-4$

We know that,

$T_n=a+(n-1)d$

$=50+(336-1)\times (-4)$

$=50+335\times (-4)$

$=50-1340$

$T_n=-1290$

Hence, this is the answer.