The 2nd, 4th and 7th terms of an AP are the first 3 consecutive terms of a GP. If the common difference of the AP is 2, find: the common ratio of the GP. the sum of the first eight terms of the GP.

Huu hapa ni utatuzi wa swali la 15:

Kwanza, tutaandika maneno ya AP na GP.
Acha $A$ iwe neno la kwanza la AP na $D$ iwe tofauti ya kawaida.
Tunapewa $D = 2$.
Maneno ya AP ni:
$a_n = A + (n-1)D$

Maneno ya $2^{nd}$, $4^{th}$ na $7^{th}$ ya AP ni:
$a_2 = A + (2-1)D = A + D = A + 2$
$a_4 = A + (4-1)D = A + 3D = A + 3(2) = A + 6$
$a_7 = A + (7-1)D = A + 6D = A + 6(2) = A + 12$

Maneno haya matatu ($a_2, a_4, a_7$) ndiyo maneno matatu ya kwanza mfululizo ya GP.
Kwa hivyo, acha $g_1 = A + 2$, $g_2 = A + 6$, na $g_3 = A + 12$.

Kwa GP, uwiano wa maneno mfululizo ni sawa:
$\frac{g_2}{g_1} = \frac{g_3}{g_2}$
Hii inamaanisha $(g_2)^2 = g_1 \cdot g_3$.

Step 1: Tafuta thamani ya $A$.
Tutabadilisha maneno ya GP kwenye fomula ya uwiano:

$$(A + 6)^2 = (A + 2)(A + 12)$$

Panua pande zote mbili:

$$A^2 + 12A + 36 = A^2 + 12A + 2A + 24$$ $$A^2 + 12A + 36 = A^2 + 14A + 24$$

Toa $A^2$ kutoka pande zote mbili:

$$12A + 36 = 14A + 24$$

Panga upya ili kutatua $A$:

$$36 - 24 = 14A - 12A$$ $$12 = 2A$$ $$A = 6$$

Kwa hivyo, neno la kwanza la AP ni 6.

Step 2: Tafuta maneno ya kwanza ya GP.
Sasa tunaweza kupata maneno ya kwanza ya GP:
$g_1 = A + 2 = 6 + 2 = 8$
$g_2 = A + 6 = 6 + 6 = 12$
$g_3 = A + 12 = 6 + 12 = 18$