Please help me solve in English plz

Hapa kuna suluhisho la swali la 1(a):

1. (a) Mfuatano wa hesabu umetungwa na nambari zinazogawanyika hasa kwa 4, huku nambari ya mwisho ikiwa 928. Nambari ya kwanza ni 4.

• Step 1: Determine the common difference. Kwa kuwa nambari zote katika mfuatano zinagawanyika hasa kwa 4, tofauti ya kawaida kati ya nambari mfululizo lazima iwe 4.

  $$d = 4$$

  Tofauti ya kawaida ni $\boxed{\text{4}}$.

• Step 2: Determine the number of terms in the series. Tunatumia formula ya nambari ya $n$-th katika mfuatano wa hesabu: $L = a + (n-1)d$. Tuna $L = 928$, $a = 4$, na $d = 4$.

  $$\begin{align*} 928 &= 4 + (n-1)4 \\ 928 - 4 &= (n-1)4 \\ 924 &= 4(n-1) \\ \frac{924}{4} &= n-1 \\ 231 &= n-1 \\ n &= 231 + 1 \\ n &= 232 \end{align*}$$

  Idadi ya nambari katika mfuatano ni $\boxed{\text{232}}$.

• Step 3: Determine the sum of all terms. Tunatumia formula ya jumla ya mfuatano wa hesabu: $S_n = \frac{n}{2}(a + L)$. Tuna $n = 232$, $a = 4$, na $L = 928$.

  $$\begin{align*} S_{232} &= \frac{232}{2}(4 + 928) \\ S_{232} &= 116(932) \\ S_{232} &= 108112 \end{align*}$$

  Jumla ya nambari zote ni $\boxed{\text{108,112}}$.

• Step 4: Determine the 80th term. Tunatumia formula ya nambari ya $n$-th: $a_n = a + (n-1)d$. Tunataka kupata nambari ya 80, kwa hivyo $n = 80$. Tuna $a = 4$ na $d = 4$.

  $$\begin{align*} a_{80} &= 4 + (80-1)4 \\ a_{80} &= 4 + (79)4 \\ a_{80} &= 4 + 316 \\ a_{80} &= 320 \end{align*}$$

  Nambari ya 80 ni $\boxed{\text{320}}$.