Question

Show that $\bf{a_{1}, a_{2},... a_{n}}$ form an AP where $\bf{a_{n}}$ is defined as below :

(i) $\bf{a_{n} = 3 + 4n }$

(ii) $\bf{a_{n} = 9 - 5n}$

Also find the sum of the first 15 terms in each case.

( Answer :- (i) 525 , (ii) (-465) )

$\bf{CBSE\:Class\:X}$
$\bf{Chapter\:5}$

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Answer 1

yes this forms an AP.

because when you subtract the 2nd term with 1st the difference is a and so on.

(1). the sum of first 15 term in the AP

an = 3 + 4n is giving by putting the values of n in the AP.

on putting n = 1 we get,

a1 = 3 +4 *1

a1 = 7.

on putting n = 15 we get,

a15 = 3 +4 *15

a15 = 3 + 60

a15 = 63

sum of AP = n/2(a + l)

here n = 15 , a = 7 and l = 63

then,

sum = 15/2( 63+7 )

sum = 15/2 * 70

sum = 15* 35

sum = 525.

(2) the sum of first 15 term in the AP.

an = 9 - 5n is given by putting the value of n as same as above.

on putting n = 1 we get,

a1 = 9 - 5*1

a1 = 9 - 5

a1 = 4

on putting n = 15 we get,

a15 = 9 - 5*15

a15 = 9 - 75

a15 = -66

sum of AP = n/2( a + l)

here n = 15, a = and l = -66

sum = 15/2 [ 4 + (-66) ]

sum = 15/2 [ 4 -66 ]

sum = 15/2 * (-62)

sum = 15 * (-31)

sum = -465.

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Answer 2

Solution:-

(i) $\bf{a_{n} = 3 + 4n }$

Taking ( n = 1).we get,

=) $\bf{a_{1}= 3 + 4×1$

=) $\bf{a_{1} = 7$

Taking ( n = 2). we get,

=) $\bf{a_{2} = 3 + 4×2$

=) $\bf{a_{2}= 11$

Taking (n=3). we get,

=) $\bf{a_{3}= 3 + 4×3$

=) $\bf{a_{3} = 15$

Now,

Common Difference = $\bf{a_{2}$ - $\bf{a_{1}$

=) C. D = 11 - 7

=) C. D = 4.

Again,

Common Difference = $\bf{a_{3}$ - $\bf{a_{2}$

=) Common Difference= 15 - 11

=) C. D. = 4

Since,

All the time the C.D is Same.

Hence, It is in A.P.

Now,

Sum of 15 Term.

=) $\bf{s_{n} = n/2[ 2a + (n-1)d]$

=) $\bf{s_{15} = 15/2[ 2×7 + ( 15 - 1)4]$

=) $\bf{s_{15} = 15/2[ 14 + 14×4]$

=) $\bf{s_{15} = 15 [ 7 + 14×2]$

=) $\bf{s_{15} = 15[ 7 + 28]$

=) $\bf{s_{15} = 525$

(ii) $\bf{a_{n} = 9 - 5n}$

Taking (n=1). we get,

=) $\bf{a_{1} = 9 - 5×1$

=) $\bf{a_{1}= 4$

Taking (n=2). we get,

=) $\bf{a_{2} = 9 - 5×2$

=) $\bf{a_{2}= -1$

Taking (n=3). we get,

=) $\bf{a_{3} = 9 - 5×3$

=) $\bf{a_{3} = -6$

Now,

Common Difference= $\bf{a_{2}$ - $\bf{a_{1}$

=) C. D = -1 - 4

=) C. D = -5

Again,

Common Difference= $\bf{a_{3}$ - $\bf{a_{2}$

=) C. D = -6 - (-1)

=) C. D = -5

Hence,

It is in A.P.

Now,

Sum of 15 Term.

=) $\bf{s_{n} = n/2[ 2a + (n-1)d]$

=)$\bf{s_{15} = 15/2 [ 2×4 + (15-1)(-5)]$

=)$\bf{s_{15} = 15/2[ 8 - 14×5]$

=) $\bf{s_{15}= 15[ 4 - 7×5]$

=) $\bf{s_{15} = 15[ -31]$

=) $\bf{s_{15} = -465$

Hence Proved!