For an arithmetic progression- 5- 112- 6- 132- - then T40 - T20 - -

The correct option is B 10 Given an Arithmetic Progressive: 5, 112, 6, 132, ..... Its first term a = 5 ,Common difference d = 112 5 = ...

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Standard XII

Mathematics

nth Term of A.P

For an arithm...

Question

For an arithmetic progression; $5, \frac{11}{2}, 6, \frac{13}{2}, . . . . .$ then $T_{40} - T_{20} =$ _________.

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Solution

The correct option is B $10$
Given an Arithmetic Progressive:
$5, \frac{11}{2}, 6, \frac{13}{2}, . . . . .$
Its first term $a = 5$ ,
Common difference $d = \frac{11}{2} - 5$
$= \frac{11 - 10}{2}$
$= \frac{1}{2}$
According to formula,
$T_{n} = a + (n - 1) d$
$∴ T_{40} = a + (40 - 1) d = a + 39 d$
$T_{20} = a + (20 - 1) d = a + 19 d$
Now, $T_{40} + T_{20} = (a + 39 d) - (a + 19 d)$
$= a + 39 d - a - 19 d$
$= 20 d$
$= 20 (\frac{1}{2})$
$= 10$ .

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For an arithmetic progression; 5,112,6,132,..... then T40−T20= _________.

The correct option is B 10Given an Arithmetic Progressive:5,112,6,132,.....Its first term a=5,Common difference d=112−5=11−102=12According to formula,Tn=a+(n−1)d∴T40=a+(40−1)d=a+39dT20=a+(20−1)d=a+19dNow, T40+T20=(a+39d)−(a+19d)=a+39d−a−19d=20d=20(12)=10.

For an arithmetic progression; $5, \frac{11}{2}, 6, \frac{13}{2}, . . . . .$ then $T_{40} - T_{20} =$ _________.