Answer:
[tex]\sum _{k=1}^4\:\:\left(3k+3\right)[/tex]
option-d
Step-by-step explanation:
we are given series as
[tex]6+9+12+15[/tex]
so, first term is 6
[tex]a_1=6[/tex]
now, we can find common difference
[tex]d=9-6[/tex]
[tex]d=3[/tex]
now, we can find kth term
[tex]a_k=a_1+(k-1)d[/tex]
we can plug values
[tex]a_k=6+(k-1)\times 3[/tex]
[tex]a_k=6+3k-3[/tex]
[tex]a_k=3k+3[/tex]
Since, there are four terms
so, k varies from 1 to 4
so, we get series as
[tex]\sum _{k=1}^4\:\:\left(3k+3\right)[/tex]
