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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.
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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
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