Question

If is an ap and a1+a4+a7+........a16 = 147, then a1+a6+a11+a/6=

Answer 1

We know that in AP the sum of first and last term is equal to second and second last term and so so
a1+a4+a7+a10+a13+a16= 147
( a1+a16)+( a4+a13)+( a7+a10)=147
( a1+a16)=( a4+a13)=( a7+a10)=147
3= 49
So asked values,
a1+a16+a11+a6= ( a1+a16)+(a11+a6)=2( a1+a16)= 2×49= 98

Answer 2

Answer:

98

Step-by-step explanation:

Given the series: $a_1+ a_4+ a_7+ ........a_{16}$ = 147 are in AP

According to the Arithemtic Progression, the sum of the first and last term is equal to the sum of the second term and second last term

                    $(a_1+ a_{16})= (a_4+ a_{13})= (a_7+ a_{10})= 147$

                    $(a_1+ a_{16})+ (a_4+ a_{13})+ (a_7+ a_{10})= 147$

                    $a_1+ a_{16}+ a_{11}+ a_6\\= (a_1+ a_{16})+ (a_{11}+ a_6)= 2(a_1+ a_{16})\\= 2\times 49= 98$