Arithmetic Progression NCERT Exercise 5.3 part four Class Ten Mathematics

Arithmetic Progression

Exercise 5.3 Part 4

Question 4: How many terms of the AP : 9, 17, 25, . . . must be taken to give a sum of 636?

Solution: Here; a = 9, d = 8 and Sn = 636

We know;

`S=n/2[2a+(n-1)d]`

Or, `636=n/2[2xx9+(n-1)8]`

Or, `636=n(9+4n-4)`

Or, `n(4n+5)=636`

Or, `4n^2+5n=636`


Or, `4n^2+5n-636=0`

Or, `4n^2-48n+53n-636=0`

Or, `4n(n-12)+53(n-12)=0`

Or, `(4n+53)(n-12)=0`

Hence, `n=(53)/(4)` and `n=12`

Taking the integral value, we have n = 12


Question 5: The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.

Solution: We know;

`S=n/2[2a+(n-1)d]`

Or, `400=n/2[2xx5+(n-1)d]`

Or, `800=n(5+45)`

Or, `50n=800`

Or, `n=800÷50=16`

Now, common difference can be calculated as follows:

`a_n = a + (n – 1)d`

Or, `45 = 5 + 15d`

Or, `15d = 40`

Or, `d = (40)/(15) = 8/3`

Thus, n = 16 and `d = 8/3`


Question 6: The first and the last terms of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?

Solution: We have; a = 17, an = 350 and d = 9

We know;

`a_n = a + (n – 1)d`

Or, `350 = 17 + (n – 1)9`

Or, `(n – 1)9 = 350 – 17`

Or, `n – 1 = 333/9 = 37`

Or, `n = 38`

Now, sum can be calculated as follows:

`S=n/2[2a+(n-1)d]`

`=(38)/(2)(17+350)`

`=19xx367=6973`



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