Question

If a=-1,d=1/2then find the 15th term of Ap and find sum of the first 25 terms

Answer 1

Given :-

a = - 1

d = ½

Required to find :-

• 15th term of the arithmetic progession ?

• Sum of the first 25 terms ?

Formula used :-

$\huge{\dagger{\boxed{\rm{ {a}_{nth} = a + ( n - 1 ) d }}}}$

$\huge{\dagger{\boxed{\rm{ {s}_{nth} = \dfrac{n}{2}[ 2a + ( n - 1 ) d ] }}}}$

Solution :-

Given that :-

a = - 1

d = ½

From the above we can conclude that ,

first term ( a ) = - 1

common difference ( d ) = ½ = 0.5

So,

Using the formula

$\huge{\dagger{\boxed{\rm{ {a}_{nth} = a + ( n - 1 ) d }}}}$

Here,

a = first term

d = common difference

n = the term number which you want to find

So, we need to find the 15th term

Here,

$\longrightarrow{\rm{ {a}_{nth} = {a}_{15}}}$

Hence,

$\longrightarrow{\tt{ {a}_{15} = - 1 + ( 15 - 1 ) \times 0.5 }}$

$\longrightarrow{\tt{ {a}_{15} = - 1 + (14) \times 0.5 }}$

$\longrightarrow{\tt{ {a}_{15} = - 1 + 7 }}$

$\longrightarrow{\tt{ {a}_{15} = 6 }}$

$\huge{\dagger{\boxed{\mathsf{ {15}^{th} \; term = 6 }}}}$

Similarly,

we need to find the sum of first 25 terms

So,

Using the formula

$\huge{\dagger{\boxed{\rm{ {s}_{nth} = \dfrac{n}{2}[ 2a + ( n - 1 ) d ] }}}}$

( symbol have their general meaning )

Here,

$\rightarrow{\rm{ {s}_{nth} = {s}_{25} }}$

So, substitute the values

$\rightarrow{\tt{ {s}_{25} = \dfrac{25}{2} [ 2(-1) + (25 - 1 )0.5 ]}}$

$\rightarrow{\tt{ {s}_{25} = 12.5 [ - 2 + ( 24 ) \times 0.5 }}$

$\rightarrow{\tt{ {s}_{25} = 12.5 [ - 2 + 12 ] }}$

$\rightarrow{\tt{ {s}_{25} = 12.5 \times 10 }}$

$\rightarrow{\tt{ {s}_{25} = 125 }}$

$\large{\dagger{\boxed{\text{ Sum of the first 25 terms = 125 }}}}$

Additional Information :-

The simplified form of the formula for finding the sum of nth terms is ;

$\large{\dagger{\boxed{\rm{ {s}_{nth} = \dfrac{n}{2}[ First\; term + Last \; term ] }}}}$

( here you should also find the 25th term of A.P. to use this formula )

Generally , this formula is not used most because it is helpful in some questions only .

So, prefer to use the complex formula .