CLASS 10 AP QUESTIO N
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SOLUTION==>>
The numbers from 1 to 500 which are multiple of 2 are:
2, 4, 6, 8.............500
This forms an AP, where
first term a = 2,
common difference = 4 - 2 = 2
last term l = 500
Toral numbers n = 500/2 = 250
Now, Sum = (n/2)*(a + l)
= (250/2)*(2 + 500)
= (250/2) * 502
= 125 * 502
= 62750
The numbers from 1 to 500 which are multiple of 5 are:
5, 10, 15, 20.............500
This forms an AP, where
first term a = 5,
common difference = 10 - 5 = 5
last term l = 500
Toral numbers n = 500/5 = 100
Now, Sum = (n/2)*(a + l)
= (100/2)*(5 + 500)
= 50 * 505
= 25250
Again, multiple of 10 are included in both i.e. in multiple of 2 and multiple of 5 also.
Now, the numbers from 1 to 500 which are multiple of 10 are:
10, 20, 30.............500
This forms an AP, where
first term a = 10,
common difference = 20 - 10 = 10
last term l = 500
Toral numbers n = 500/10 = 50
Now, Sum = (n/2)*(a + l)
= (50/2)*(10 + 500)
= 25 * 510
= 12750
Now, the sum of integers from 1 to 500 which are multiple of 2 or 5 = sum of multiple of 2 + sum of multiple of 5 - sum of multiple of 2 and 5
= 62750 + 25250 - 12750
= 88000 - 12750
= 75250
iii) We first find the LCM of 2 and 5 which is 10.
Now all those integers which are multiples of 10 are also the multiples of 2 and 5.
Therefore, multiples of 2 as well as of 5 between 1 and 500 are:
10, 20, 30, ...., 490
Series forms an AP with first term,a=10 and common difference,d=20-10=10.
Let total number of terms of this AP be n.
Therefore, nth term of AP, an = Last term, l= 490
an=a(n-1)d=l;
10+(n-1)10=490;
(n-1)10=480;
n-1=48;
n=48+1=49;
n=49
Thus,sum of n terms of AP is given as:
S₄₉=49/2 (10+490);
=49/2 (500);
=49×250=12250
3) A8=a+7d=a+d/2 2(a+7d)=a+d 2a+14d= a+d 2a-a+14d-d=0 a+13d=0. ...... (i)
A+10d=a+3d/3+1 A+10d=(a+3d+3)/3 3(a+10d)=a+3d+3 3a-a+30d-3d-3=0 2a+27d=3. ... (ii) Solve (i)&(ii) Eq(i) multiply by 2 2a+26d=0......(iii) 2a+27d=3.......(iv)
Solve (iii) & (iv) D=3
Put d in (i) A+39=0 A= -39**** D=3
A15=a+14d =. - 39+14(3)=. - 39+42=. 3
A15 is 3
Answer is 3
4)Let the first term and the common difference of the A.P are a and d respectively.
Since the A.P contains 37 terms. So, the middle most term is (37+1)/2 th term = 19th term.
Thus, three middle most terms of this A.P.are 18th, 19th and 20th terms.
Given a 18 + a 19 + a 20 = 225
⇒ (a + 17d) + (a + 18d) + (a + 19d) = 225
⇒ 3(a + 18d) = 225
⇒ a + 18d = 75
⇒ a = 75 – 18d … (1)
According to given information
a 35 + a 36 + a 37 = 429
⇒ (a + 34d) + (a + 35d) + (a + 36d) = 429
⇒ 3(a + 35d) = 429
⇒ (75 – 18d) + 35d = 143
⇒ 17d = 143 – 75 = 68
⇒ d = 4
Substituting the value of d in equation (1), it is obtained
a = 75 – 18 × 4 = 3
Thus, the A.P. is 3, 7, 11, 15 …
Hope! This will help you.
Cheers!