Verified Answer :

dalendrk

\(-3+(-1)+1+...\\\\a_1=-3;\ a_2=-1;\ a_3=1\\\\It's\ an\ arithmetic\ sequence\ where\ the\ common\ difference\\is\ equal\ d=a_2-a_1=-1-(-3)=-1+3=2.\\\\a_n=a_1+(n-1)d\\\\a_n=-3+(n-1)\cdot2=-3+2n-2=2n-5\\\\Answer:\boxed{\sum\limits_{n=1}^7(2n-5)}\to\fbox{a.}\)

JeanaShupp

Answer: \(\sum^{7}_{n=1}(2n-5)\)

Step-by-step explanation:

Given: The first term of the series \(a= -3\)

The second term of the series \(a_2= -1\)

The third term of the series \(a_3= 1\)

The common difference=\(d=a_3-_a_2=a_2-a_1=2\)

Now, the nth term of this series will be given by :-

\(a_n=a+d(n-1)\\\\\Rightarrow\ a_n=-3+2(n-1)\\\\\Rightarrow\ a_n=-3+2n-2\\\\\Rightarrow\ a_n=2n-5\)

Hence, the expression defines the given series for seven terms will be

\(\sum^{7}_{n=1}(2n-5)\)

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