Sum of an Arithmetic Sequence II

$a_{50}$ is the "middle" term, and hence also the mean, of both arithmetic progressions

$S = a_{2} + a_{5} + a_{8} + .... + a_{98}$ and $T = a_{1} + a_{2} + a_{3} + .... + a_{99}.$

As there are $33$ terms in $S$ and $99$ in $T,$ we then know that

$S = 33a_{50}, T = 99a_{50} \Longrightarrow T = 3S = 3 \times 205 = \boxed{615}.$

Considering that (A(n-1)+ A(n+1)) /2 = A(n) We can conclude that (a1+a3)/2 =a2 (a4 +a6)/2= a5 So, all the terms that are missing are equal to twice the sum of (a2+a5+a8...). So, 2*205 +205 = 615.

We can break the series into three sub-series: $\{a_1, a_4, \cdots,a_{97} \}$ , $\{a_2, a_5, \cdots,a_{98} \}$ and $\{a_3, a_6, \cdots,a_{99} \}$ . Notice that all sub-series have 33 terms. To get the sum of the first sub-series, we need to subtract $33d$ from the sum of second sub-series, which is $205$ and to get the sum of third series, we should add $33d$ to $205$ . The sum $\sum_{i=1}^{99} a_i = 205 - 33d + 205 + 205 + 33d = 615$ .

Let $S=a_2+a_5+a_8+\ldots+a_{98}$ . The sum of a arithmetic progression is defined by the sum of the first and the last elements multiplied by the number of elements divided by two. So

$\begin{aligned} S&=\dfrac{(a_2+a_{98})\cdot33}{2}\\ &=\dfrac{(a_1+r+a_1+97r)\cdot33}{2}\\ &=\dfrac{(2a_1+98r)\cdot33}{2}=205 \end{aligned}$

Let $T=\displaystyle \sum_{k=1}^{99} a_k$ . So

$\begin{aligned} T&=\dfrac{(a_1+a_{99})\cdot99}{2}\\ &=\dfrac{(a_1+a_1+98r)\cdot99}{2}\\ &=3\cdot\dfrac{(2a_1+98r)\cdot33}{2}\\ &=3\cdot S\\ &=3\cdot205=615 \end{aligned}$

$For\:this\:arithmetic\:progression,$ $\begin{aligned} a_{ k } & = & a_{ 1 }+(k-1)x \\ & & \\ \sum _{ k=1 }^{ 99 }{ a_{ k } } & = & \sum _{ k=1 }^{ 99 }{ a_{ 1 }+(k-1)x } \\ & = & 4851x+99a_{ 1 }\qquad \qquad (1) \\ & & \\ a_{ 2 }+a_{ 5 }+a_{ 8 }+\dots +a_{ 98 } & = & \sum _{ k=1 }^{ 33 }{ a_{ 3k-1 } } \\ & = & \sum _{ k=1 }^{ 33 }{ a_{ 1 }+(3k-2)x } \\ & = & 1617x+33a_{ 1 }\qquad \qquad (2) \end{aligned}$

$It\:can\:be\:observed\:that\:eqn.\:(1)\:is\:three\:times\:eqn.\:(2)$ $\begin{aligned} 4851x+99a_{ 1 } & = & 3(1617x+33a_{ 1 }) \\ \sum _{ k=1 }^{ 99 }{ a_{ k } } & = & 3(205) \\ \sum _{ k=1 }^{ 99 }{ a_{ k } } & = & 615 \end{aligned}$

$\frac{2+1}{3} ,\frac{5+1}{3}, \frac{8+1}{3}, \ldots , \frac{98+1}{3} = 1, 2, 3, \ldots , 33$ . There are 33 terms in the series shown in the problem. Therefore,

$\dfrac{33(a_2+a_{98})}{2} =205$

$(a_2+a_{98}) = \dfrac{410}{33}$

There are a total of 99 terms in the second series, so

$\dfrac{99 \times \frac{410}{33}}{2} = 615$

I feel the problem is a little over rated.. The solution is very simple.. From the concept of AM ...for successive terms of the AP we have a1+a3=2(a2); so adding a2 on both sides will give the proper sequence on LHS...and RHS is 3(a2)...similarly we can observe that it follows the same for a5 ,a8, and so on... Mulitplying the given result with 3 gives the required answer.i.e 3(205)=615

the number of terms: $\frac{98-2}{3}+1 = 33$ .

the difference of the terms is $a_{5}-a_{1} =(a+4b)-(a+b) = 3b$

so the total sum is $\frac{n}{2}(a + U_{n})$ where $U_{n}$ is the last term and $n$ is the number of terms.

Sum = $\frac{33}{2}((a+b)+(a+97b)) = 205$ $\Leftrightarrow a + 49b = \frac{205}{33} ... (*)$ .

Now, let's get into the real sum. $\sum_{k=1}^{99}a_{k} = \frac{99}{2}(a+(a+98b) = 99 * (a+49b)$ . By substituting (*), we get:

$\sum_{k=1}^{99}a_{k} = \frac{99 * 205}{33} = \boxed{615}$

The general sum=99/2(a1+a99) We have 205= 33/2( a2 + a98 ) So 33/2[ (a1+d) +(a99 - d ) ]= 205 Then 33/2(a1 + a99 )= 205 =1/3(the general sum) So the answer =3* 205= 615