Question

There 50 terms in an ap the sum of first 10 terms of an ap is210 the sum of last 15 terms is 2565 find the ap​

Answer 1

Solution :

$\bf{\blue{\underline{\bf{Given\::}}}}$

There 50 terms in an A.P. the sum of first 10 terms of an A.P. is 210,the sum of last 15 term is 2565.

$\bf{\blue{\underline{\bf{To\:find\::}}}}$

The arithmetic progression.

$\bf{\blue{\underline{\bf{Explanation\::}}}}$

We know that formula of the sum of an A.P;

$\boxed{\bf{S_n=\frac{n}{2} [2a+(n-1)d]}}}}}$

• a is the first term
• d is the common difference.

A/q

$\longrightarrow\sf{S_{10}=\cancel{\dfrac{10}{2}} \bigg[2a+(10-1)d\bigg]}\\\\\\\longrightarrow\sf{210=5\big[2a+9d\big]}\\\\\\\longrightarrow\sf{\cancel{\dfrac{210}{5} }=2a+9d}\\\\\\\longrightarrow\sf{42=2a+9d........................(1)}$

&

Sum of last 15th term :

$\longrightarrow\sf{S_{50}-S_{35}=2565}\\\\\\\longrightarrow\sf{\cancel{\dfrac{50}{2}} \bigg[2a+(50-1)d\bigg]-\dfrac{35}{2} \bigg[2a+(35-1)d\bigg]=2565}\\\\\\\longrightarrow\sf{25\big[2a+49d\big]-\dfrac{35}{\cancel{2}} \bigg[\cancel{2}a+\cancel{34}d\bigg]=2565}\\\\\\\longrightarrow\sf{50a+1225d-35[a+17d]=2565}\\\\\\\longrightarrow\sf{50a+1225d-35a-595d=2565}\\\\\\\longrightarrow\sf{50a-35a+1225d-595d=2565}\\\\\\\longrightarrow\sf{15a+630d=2565}\\\\\\\longrightarrow\sf{15(a+42d)=2565}$

$\longrightarrow\sf{a+42d=\cancel{\dfrac{2565}{15} }}\\\\\\\longrightarrow\sf{a+42d=171............................(2)}$

$\underline{\underline{\bf{\pink{Using\:Substitution\:method\::}}}}$

From equation (2),we get;

$\mapsto\sf{a+42d=171}\\\\\mapsto\sf{a=171-42d..................................(3)}$

Putting the value of a in equation (1),we get;

$\longrightarrow\sf{2(171-42d)+9d=42}\\\\\longrightarrow\sf{342-84d+9d=42}\\\\\longrightarrow\sf{342-75d=42}\\\\\longrightarrow\sf{-75d=42-342}\\\\\longrightarrow\sf{-75d=-300}\\\\\longrightarrow\sf{d=\cancel{\dfrac{-300}{-75} }}\\\\\\\longrightarrow\sf{\red{d=4}}$

Putting the value of d in equation (3),we get;

$\longrightarrow\sf{a=171-42(4)}\\\\\longrightarrow\sf{a=171-168}\\\\\longrightarrow\sf{\red{a=3}}$

Now;

$\underline{\underline{\sf{\pink{Arithmetic\:progression\:(A.P)\::}}}}$

We know that;

$\boxed{\bf{a_{n}=a+(n-1)d}}}}$

$\bullet\sf{a_{1}=3+(1-1)4}\\\\\implies\sf{3+0\times 4}\\\\\implies\sf{3+0}\\\\\implies\boxed{\bf{\red{3}}}}}}$

$\bullet\sf{a_{2}=3+(2-1)4}\\\\\implies\sf{3+1\times 4}\\\\\implies\sf{3+4}\\\\\implies\boxed{\bf{\red{7}}}}$

$\bullet\sf{a_{3}=3+(3-1)4}\\\\\implies\sf{3+2\times 4}\\\\\implies\sf{3+8}\\\\\implies\boxed{\bf{\red{11}}}}$

Thus;

The A.P. is 3, 7, 11...............

Answer 2

GIVEN:

Sum of first 10 terms of an AP = 210

Sum of last 15 terms of AP = 2565

TO FIND:

The arithmetic progression.

SOLUTION:

Sum of first 10 terms of AP = 210

We know that,

Sum of terms in an AP = n/2[2a + (n - 1)d]

10/2 [2a + 9d] = 210

[ 2a + 9d ] = 210/5

2a + 9d = 42 -----(1)

Sum of last 15 terms of AP = 2565

[ Sum of last 15 terms = Sum of 50 terms - sum of first 35 terms]

2565 = 50/2 [2a + 49d] - 35/2[ 2a + 34d]

2565 = 25[2a + 49d] - 35/2[ 2a + 34d]

2565 = 50a + 1225 - 35d - 595d

2565 = 15a + 630d

Divide the whole equation with 15

171 = a + 42d -----(2)

Solve both the Equations,

Multiply the second equation with 2

2a + 9d = 42

a + 42d = 171 × 2

------------------

2a + 9d = 42

2a + 84d = 342

--------------------

-75d = -300

d = 300/75

d = 4

Common Difference = 4

Substitute the (d) value in eq - (1)

2a + 9(4) = 42

2a + 36 = 42

2a = 42 - 36

==> a = 6/2

==> a = 3

First term = 3

Second term = 3 + 4 = 7

Third term = 7 + 4 = 11

Therefore, the AP is 3, 7, 11, 15,......