Calculation is beautiful

Let $a_2+a_4+a_6+\ldots+a_{98}=S$

$2S= a_2+a_2+a_4+a_4+\ldots +a_{98}+a_{98}$

$2S= a_1+d+a_2+a_3+d+a_4+\ldots +a_{97}+d+a_{98}$

$=a_1+a_2+a_3+a_4+\ldots+a_{97}+a_{98}+49$ , since $d=1$

Therefore, $2S=137+49=186 \Rightarrow S=\boxed{93}$

Given: $common-difference(d)=1$ , $first-term=a_1=a(say)$

$\therefore a_1+a_2+.......+a_{98}=137$

$\dfrac{98}{2}.\left[ 2a+97 \right] =137$

$\implies 2a+97=\dfrac{137}{49}$ /

To find : $a_2+a_4+......+a_{98} \left( 49 \ terms \right)$

$\quad \quad =\dfrac{49}{2} \left[ a_2+a_{98} \right]$

$\quad \quad =\dfrac{49}{2} \left[ (a+1)+(a+97) \right]$

$\quad \quad =\dfrac{49}{2} \left[ 2a+97+1 \right]$

$\quad \quad =\dfrac{49}{2} \left[ \dfrac{137}{49}+1 \right]$

$\quad \quad =\dfrac{137}{2}+\dfrac{49}{2}=\dfrac{186}{2}=\boxed{93}$

$Let\quad { a }_{ 1 }=x\\ Since\quad { a }_{ 1 },{ \quad a }_{ 2 },\quad { a }_{ 3 },...\quad is\quad an\quad arithmetic\quad progression\\ with\quad common\quad difference\quad 1,\\ { a }_{ 1 }+{ a }_{ 2 }+{ a }_{ 3 }+...+{ a }_{ 98 }=137\\ x+(x+1)+(x+2)+(x+3)+...+(x+97)=137\\ 98x+\frac { 97(98) }{ 2 } =137\\ 98x+97(49)=137\\ 98x+4753=137\\ 98x=-4616\\ x=-\frac { 2308 }{ 49 } \\ \\ { a }_{ 2 }+{ a }_{ 4 }+{ a }_{ 6 }+...+{ a }_{ 98 }\\ =(x+1)+(x+3)+(x+5)+...+(x+97)\\ =49x+\frac { 49(1+97) }{ 2 } =49x+49(49)\\ =49x+2401\\ \\ Since\quad x=-\frac { 2308 }{ 49 } ,\\ 49\left( -\frac { 2308 }{ 49 } \right) +2401\\ =-2308+2401=93\\ \\ \boxed { { \therefore \quad a }_{ 2 }+{ a }_{ 4 }+{ a }_{ 6 }+...+{ a }_{ 98 }=93 } \\$

Given, $a_{1} +a_{2} +a_{3}+ . . . +a_{98} =137$

$=>a_{1} +(a_{1} +1) +(a_{1} +2)+ . . .+(a_{1}+97) =137$

$=> 98 a_{1} + 1+2+ . . . +97 =137$

$=> 98 a_{1} + (97*98)/2 =137$

$=>a_{1} = -2308/49$

So, $a_{2} = -2308/49 +1 =-2259/49$

Assume, $a_{2} + a_{4} + . . . +a_{98} = S$

Number of terms in this arithmetic progression is =49.

$S = 49/2 (2(-2259/49) + 48*2)$

$=-2259 + 2352$

$= \boxed{93}$

first, the Am of the series (median) $AM=\dfrac{a_1 +a_2 \dots\dots\dots +a_{98}}{98} =\dfrac{a_{48}+a_{49}}{2}=\dfrac{137}{98}$ $AM =\dfrac{a_2 +a_4 +a_6 +\dots\dots+ a_{98}}{49}=\dfrac{a_{48}+a_{50}}{2}$ $a_{50}=a_{49}+1,AM= \dfrac{a_{48}+a_{49}+1}{2}=\dfrac{1}{2}+\dfrac{a_{48}+a_{49}}{2}$ $AM= \dfrac{1}{2}+\dfrac{137}{98}=\dfrac{93}{49}$ $\dfrac{a_2 +a_4 +a_6 +\dots\dots +a_{98}}{49}=\dfrac{93}{49}$ $a_2 +a_4 +a_6 +\dots\dots +a_{98}=\boxed{93}$

There are 98 terms in the series A1, A2... A98, which is 49 pairs

If you consider each pair of consecutive numbers, you have A1, A1+1, A3, A3+1..., which is equal to our series, and its sum still equals 137.

If you simplify that series it would be 2 * sum of odd numbers + 49 = 137 or 2 sum of odd numbers = 88 Sum of odd numbers - 44

We need the even numbers = sum of odd + 49 (since each of the 49 even terms are one unit larger) = 44 + 49 = 93. OR you could say the sum of the even numbers is 137 (sum of full series) minus the sum of odd numbers (44): 137-44 = 93. Take your pick how you want to think about it.

$a_{98}=a_1+97$

$S=$ $\frac{n}{2}$ $(a_1+a_n)$

$137=$ $\frac{98}{2}$ $(a_1+a_{98})$

$\frac{137}{49}$ $=$ $a_1+a_{98}$

$but:$ $a_{98}=a_1+97$

$\frac{137}{49}$ $=$ $a_1+$ $a_1+97$

$a_1=$ $\frac{-2308}{49}$

$a_{98}=$ $\frac{2445}{49}$

The number of terms of the second Arithmetic Progression is $n=$ $\frac{98}{2}$ $=49.$ The first term of the second Arithmetic Progression is the second term of the first Arithmetic Progression, $a_1=$ $\frac{-2308}{49}$ $+1=$ $\frac{-2259}{49}$ . The sum of the second Arithmetic Progression is

$S=$ $\frac{n}{2}$ $(a_1+a_n)$

$S=$ $\frac{49}{2}$ $(\frac{-2259}{49}+\frac{2445}{49})$ $=$ $93$

Given: a1+a2+a3+...+a96+a97+a98 = a1+(a1+1)+(a1+2)+(a1+3)+...+(a1+96)+(a1+97)=137

Problem: a2+a4+a6+...+a94+a96+a98

Sol'n: 98a1+(1+2+3+...+95+96+97)=137 --> 98a1+4753=137 --> a1 = -2308/49

a2+a4+a6+...+a94+a96+a98 = 49a1+(1+3+5+...+93+95+97) --> 49a1+2401

plug in the value for a1 to 49a1+2401 then we get 93

Add 1-6: 1 + 2 + 3 + 4 + 5 + 6 = 21

Add even terms: 2 + 4 + 6 = 12

Add 1-8: 21+7+8= 36

Add even terms = 20

Add 1-10: 36+9+10=55

Add even terms: = 20+10 = 30

The sum of n consecutive numbers plus 1/2n, all divided by 2 is the sum of the even terms.

So 98 consecutive terms totals 137 means (137 + 49)/2 should be the sum of the even terms, which is 93.