This question will definitely lead to a simultaneous equation. Kindly observe:
The sum of an nth term= Sn= ½(n)(2a + (n - 1)d), where a= first term and d= common difference in an Arithmetic Progression(AP).
Now, the sum of the first six terms= S6= ½(6)(2a + (6- 1)d)= 9
= 3(2a + 5d)= 9
= 3(2a) + 3(5d)= 9
= 6a + 15d= 9
Divide all sides by three and we will have:
2a + 5d= 3 •••> (1)
Another, the fourth term is 2.
Recall, Tn= a + (n - 1)d
So, T4= a + (4 - 1)d= 2
= a + 3d= 2 •••>(2)
2a + 5d= 3 •••>(1)
a + 3d= 2 •••>(2)
-------------------------- Using the elimination method,
2a + 5d= 3 •••>(1) × 1
a + 3d= 2 •••>(2) × - 2
---------------------------
2a + 5d= 3 •••>(3)
-2a - 6d= - 4 •••>(4)
-------------------- Add up equation (3) and (4)
- d= - 1
d= 1
From equation (2),
a + 3d= 2 (d= 1)
a + 3(1)= 2
a + 3= 2
a= 2 - 3
a= -1
Thus, a= -1; d= 1
The seventh term of the AP= T7= a + 6d
= (-1) + 6(1)
= -1 + 6
= 5
3 weeks ago