The value of $x + y + z$ is $15$ , if $a, x, y, z, b$ are in A.P. Then find $a + b$

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Solution

Let A.P. have first term $a$ and common difference $d$
$\Rightarrow x = a + d, y = a + 2 d, y = a + 3 d, b = a + 4 d$
We have, $x + y + z = 15 \Rightarrow 3 a + 6 d = 15 \Rightarrow a + 2 d = 5 \Rightarrow (1)$
Hence $a + b = a + (a + 4 d) = 2 a + 4 d = 2 (a + 2 d) = 2 (5) = 10$ using (1)

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The value of x+y+z is 15, if a,x,y,z,b are in A.P. Then find a+b

Let A.P. have first term a and common difference d⇒x=a+d,y=a+2d,y=a+3d,b=a+4dWe have, x+y+z=15⇒3a+6d=15⇒a+2d=5⇒(1)Hence a+b=a+(a+4d)=2a+4d=2(a+2d)=2(5)=10 using (1)

The value of $x + y + z$ is $15$ , if $a, x, y, z, b$ are in A.P. Then find $a + b$

Let A.P. have first term $a$ and common difference $d$
$\Rightarrow x = a + d, y = a + 2 d, y = a + 3 d, b = a + 4 d$
We have, $x + y + z = 15 \Rightarrow 3 a + 6 d = 15 \Rightarrow a + 2 d = 5 \Rightarrow (1)$
Hence $a + b = a + (a + 4 d) = 2 a + 4 d = 2 (a + 2 d) = 2 (5) = 10$ using (1)