Advertisements
Advertisements
Question
In a Arithmetic Progression (A.P.) the fourth and sixth terms are 8 and 14 respectively. Find that:
(i) first term
(ii) common difference
(iii) sum of the first 20 terms.
Advertisements
Solution 1
Let the first term of the sequence is a and the common difference is d.
a4 = a +3d = 8 ...(1)
a6 = a + 5d = 14 ...(2)
- - -
_______________________________
-2d = - 6
d = 3
Put d = 3 in equation (1)
a + 3 × 3 = 8
a = - 1
∴ (i) First term (a) = –1
(ii) Common difference (d) = 3
(iii) Sum of the first 20 terms = Sn `= n/2 [2a + (n-1)d]`
` = 20/2 [2 xx (-1) + 19 xx 3]`
` = 20/2 [-2+57]`
= 10 × 55 = 550
Solution 2
Let the first term of the sequence is a and the common difference is d.
a4 = a +3d = 8 ...(1)
a6 = a + 5d = 14 ...(2)
- - -
_______________________________
-2d = - 6
d = 3
Put d = 3 in equation (1)
a + 3 × 3 = 8
a = - 1
∴ (i) First term (a) = –1
(ii) Common difference (d) = 3
(iii) Sum of the first 20 terms = Sn `= n/2 [2a + (n-1)d]`
` = 20/2 [2 xx (-1) + 19 xx 3]`
` = 20/2 [-2+57]`
= 10 × 55 = 550
Solution 3
Let the first term of the sequence is a and the common difference is d.
a4 = a +3d = 8 ...(1)
a6 = a + 5d = 14 ...(2)
- - -
_______________________________
-2d = - 6
d = 3
Put d = 3 in equation (1)
a + 3 × 3 = 8
a = - 1
∴ (i) First term (a) = –1
(ii) Common difference (d) = 3
(iii) Sum of the first 20 terms = Sn `= n/2 [2a + (n-1)d]`
` = 20/2 [2 xx (-1) + 19 xx 3]`
` = 20/2 [-2+57]`
= 10 × 55 = 550
Solution 4
Let the first term of the sequence is a and the common difference is d.
a4 = a +3d = 8 ...(1)
a6 = a + 5d = 14 ...(2)
- - -
_______________________________
-2d = - 6
d = 3
Put d = 3 in equation (1)
a + 3 × 3 = 8
a = - 1
∴ (i) First term (a) = –1
(ii) Common difference (d) = 3
(iii) Sum of the first 20 terms = Sn `= n/2 [2a + (n-1)d]`
` = 20/2 [2 xx (-1) + 19 xx 3]`
` = 20/2 [-2+57]`
= 10 × 55 = 550
Solution 5
Let the first term of the sequence is a and the common difference is d.
a4 = a +3d = 8 ...(1)
a6 = a + 5d = 14 ...(2)
- - -
_______________________________
-2d = - 6
d = 3
Put d = 3 in equation (1)
a + 3 × 3 = 8
a = - 1
∴ (i) First term (a) = –1
(ii) Common difference (d) = 3
(iii) Sum of the first 20 terms = Sn `= n/2 [2a + (n-1)d]`
` = 20/2 [2 xx (-1) + 19 xx 3]`
` = 20/2 [-2+57]`
= 10 × 55 = 550
APPEARS IN
RELATED QUESTIONS
A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A of radii 0.5, 1.0 cm, 1.5 cm, 2.0 cm, .... as shown in figure. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take `pi = 22/7`)
[Hint: Length of successive semicircles is l1, l2, l3, l4, ... with centres at A, B, A, B, ... respectively.]
If numbers n – 2, 4n – 1 and 5n + 2 are in A.P., find the value of n and its next two terms.
Find the sum of first 12 natural numbers each of which is a multiple of 7.
What is the sum of first n terms of the AP a, 3a, 5a, ....
In an A.P. the 10th term is 46 sum of the 5th and 7th term is 52. Find the A.P.
Write the value of x for which 2x, x + 10 and 3x + 2 are in A.P.
If four numbers in A.P. are such that their sum is 50 and the greatest number is 4 times, the least, then the numbers are
If an = 3 – 4n, show that a1, a2, a3,... form an AP. Also find S20.
If the sum of the first m terms of an AP is n and the sum of its n terms is m, then the sum of its (m + n) terms will be ______.
The nth term of an A.P. is 6n + 4. The sum of its first 2 terms is ______.
